 # SOA Sample Exam Solutions [ ]

```SOA Sample Exam MFE/3F
Solutions
SOA Sample Exam Solutions
Solution 1
A
Chapter 1, Put-Call Parity
We can use put-call parity to solve this problem:
CEur ( K ,T ) + Ke - rT = S0 + PEur ( K ,T )
[CEur ( K ,T ) - PEur ( K ,T )] - S0 = - Ke -rT
0.15 - 60 = -70e -4r
-59.85 = -70e -4r
Ê 59.85 ˆ
ln Á
= -4r
Ë 70 ˜¯
r = 0.03916
Solution 2
D
Chapter 2, Arbitrage
Let X be the number of calls with a strike price of \$55 that are purchased for Mary’s
portfolio. If we assume that the net cost of establishing the portfolio is zero, then we can
solve for X:
11 - 3 ¥ 6 + 1 + 3 X = 0
X =2
The table below shows that regardless of the stock price at time T, Mary’s payoff is
positive. Therefore, Mary is correct. This implies that John is incorrect.
Mary’s Portfolio
Time T
Transaction
Time 0
ST < 40
40 £ ST £ 50
50 £ ST £ 55
55 < ST
–11.00
0.00
ST - 40
ST - 40
ST - 40
Sell 3 of C (50)
3(6.00)
0.00
0.00
-3(ST - 50)
-3(ST - 50)
–2(3.00)
0.00
0.00
0.00
2(ST - 55)
-1.00
erT
erT
erT
erT
0.00
erT
erT + ST - 40
erT + 110 - 2ST
erT
Lend \$1
Total
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SOA Sample Exam MFE/3F
Solutions
Let Y be the number of calls and puts with a strike price of \$50 that are sold for Peter. If
we assume that the net cost of establishing the portfolio is zero, then we can solve for Y:
2 ¥ 3 - 2 ¥ 11 + 11 - 3 + 2 - Y ¥ 6 + Y ¥ 8 = 0
2Y = 6
Y =3
In evaluating Peter’s portfolio, we can make use of the fact that purchasing a call option
and selling a put option is equivalent to purchasing a prepaid forward on the stock and
borrowing the present value of the strike price. We can see this by writing put-call parity
as:
CEur ( K ,T ) - PEur ( K ,T ) = F0,PT ( S ) - Ke - rT
Therefore, purchasing a call option and selling a put option results in a payoff of:
ST - K
Since Peter purchases offsetting amounts of puts and calls for any given strike price, we
can use this result to evaluate his payoffs.
Peter’s Portfolio
Transaction
Time 0
Buy 2 of C (55) 2(11.00 - 3.00)
& sell 2 of P(55)
Buy 1 of C(40) 3.00 - 11.00
& sell 1 of P(40)
Sell 3 of C (50) 3(6.00 - 8.00)
Lend \$2
Total
Time T
2(ST - 55)
ST - 40
-3(ST - 50)
-2.00
2erT
0.00
2erT
Peter’s portfolio is certain to have a positive payoff at time T, so Peter is correct.
Solution 3
13.202%
Chapter 2, Application of Option Pricing Concepts
We are told that the price of a European put option with a strike price of \$103 has a value
of \$15.21. The payoff of the put option at time 1 is:
Max [0,103 - S (1)]
If we can express the payoff of the single premium deferred annuity in terms of the
expression above, then we will be able to obtain the price of the annuity.
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SOA Sample Exam MFE/3F
Solutions
The payoff at time 1 is:
È S (1)
˘
Time 1 Payoff = p (1 - y%) Max Í
,1.03˙
100
Î
˚
1
= p (1 - y%) ¥
¥ Max [S (1),103]
100
1
= p (1 - y%) ¥
¥ Max [0,103 - S (1)] + S (1)
100
(
)
The current value of a payoff of Max [0,103 - S (1)] at time 1 is \$15.21, and the current
value of a payoff of S (1) at time 1 is 100. Therefore, the current value of the payoff is:
Current value of payoff = p (1 - y%) ¥
1
(15.21 + 100)
100
For the company to break even on the contract, the current value of the payoff must be
equal to the single premium of p :
1
(15.21 + 100) = p
100
1
(1 - y%) ¥
(15.21 + 100) = 1
100
y% = 0.13202
p (1 - y%) ¥
Solution 4
C
Chapter 4, Two-Period Binomial Model
Since the stock does not pay dividends, the price of the American call option is equal to
the price of an otherwise equivalent European call option.
The stock price tree and the associated option payoffs at the end of 2 years are:
32.9731
Call Payoff
10.9731
22.1028
0.1028
14.8161
0.0000
25.6800
20.0000
17.2140
The risk-neutral probability of an upward movement is:
p* =
e( r -d )h - d e(0.05 - 0.00)1 - 0.8607
=
= 0.4502
u-d
1.2840 - 0.8607
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SOA Sample Exam MFE/3F
Solutions
The value of the call option is:
V (S0 , K ,0)  e r( hn )
n
 n 

  i  ( p*)(n i) (1  p*)i V (S0un idi , K , hn)
i 0 

 e 0.05(2) (0.4502)2 (10.9731)  2(0.4502)(1  0.4502)(0.1028)  0 


 2.0585
Solution 5
Chapter 4, Three-Period Binomial Model for Currency
0.2255
The values of u and d are:
u=e
( r - rf )h +s h
= e(0.08 - 0.09)0.25 + 0.30 0.25 = 1.15893
( r - rf )h -s h
= e(0.08 - 0.09)0.25 - 0.30 0.25 = 0.85856
d=e
The risk-neutral probability of an upward movement is:
p* 
e
( r  rf )h
 d e(0.08  0.09)(0.25)  0.85856

 0.46257
ud
1.15893  0.85856
The tree of pound prices and the tree of option prices are below:
Pound
1.4300
1.9207
1.6573
1.2277
1.4229
1.0541
2.2259
1.6490
1.2216
0.9050
American Put
0.2255
0.0939
0.3473
0.0000
0.1783
0.5059
0.0000
0.0000
0.3384
0.6550
The tree of prices for the American put option is found by working from right to left. The
rightmost column is found as follows:
Max [0, 1.5600 - 2.2259] = 0.0000
Max [0, 1.5600 - 1.6490] = 0.0000
Max [0, 1.5600 - 1.2216] = 0.3384
Max [0, 1.5600 - 0.9050] = 0.6550
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SOA Sample Exam MFE/3F
Solutions
The prices after 6 months are found using the risk-neutral probabilities. If the exchange
rate falls to 1.0541 at the end of 6 months, then early exercise is optimal:
{
}
Max e -0.08(0.25) [(0.46257)(0.0000) + (1 - 0.46257)(0.0000)] ,1.5600 - 1.9207
{
= Max {0.0000, -0.3607} = 0.0000
}
Max e -0.08(0.25) [(0.46257)(0.0000) + (1 - 0.46257)(0.3384)] ,1.5600 - 1.4229
{
= Max {0.1783,0.1371} = 0.1783
}
Max e -0.08(0.25) [(0.46257)(0.3384) + (1 - 0.46257)(0.6550)] ,1.5600 - 1.0541
= Max {0.4985,0.5059} = 0.5059
The prices after 3 months are:
{
}
Max e -0.08(0.25) [(0.46257)(0.0000) + (1 - 0.46257)(0.1783)] ,1.5600 - 1.6573
{
= Max {0.0939, -0.0973} = 0.0939
}
Max e -0.08(0.25) [(0.46257)(0.1783) + (1 - 0.46257)(0.5059)] ,1.5600 - 1.2277
= Max {0.3473,0.3323} = 0.3473
The current price of the option is:
{
}
Max e -0.08(0.25) [(0.46257)(0.0939) + (1 - 0.46257)(0.3473)] ,1.5600 - 1.4300
= Max {0.2255,0.1300} = 0.2255
The value of the American put option at time 0 is 0.2255.
Solution 6
C
Chapter 7, Black-Scholes Call Price
The first step is to calculate d1 and d2 :
d1 =
ln( S / K ) + (r - d + 0.5s 2 )T
s T
=
ln(20 / 25) + (0.05 - 0.03 + 0.5 ¥ 0.242 ) ¥ 0.25
0.24 0.25
= -1.75786
d2 = d1 - s T = -1.75786 - 0.24 0.25 = -1.87786
We have:
N (d1 )  N ( 1.75786)  0.03939
N (d2 )  N ( 1.87786)  0.03020
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SOA Sample Exam MFE/3F
Solutions
The value of one European call option is:
CEur  Se  T N (d1 )  Ke rT N (d2 )
 20e 0.03(0.25)  0.03939  25e 0.05(0.25)  0.03020  0.03629
The value of 100 of the European call options is:
100 ¥ 0.03629 = 3.629
Solution 7
\$7.62 million Chapter 7, Currency Options and Black-Scholes
We can draw a couple of conclusions from Statement (vi):

Since the logarithm of the dollar per yen exchange rate is an arithmetic Brownian
motion, the dollar per yen exchange rate follows geometric Brownian motion.

Since the dollar per yen exchange rate follows geometric Brownian motion, the
Black-Scholes framework applies. This means that put and call options on yen can
be priced using the Black-Scholes formula.
Company A has decided to buy a dollar-denominated put option with yen as the
underlying asset. The domestic currency is dollars, and the current value of the one yen
is:
1
= 0.0083333 dollars
120
x0 =
Since the option is at-the-money, the strike price is equal to the value of one yen:
K=
1
= 0.0083333 dollars
120
The domestic interest rate is 3.5%, and the foreign interest rate is 1.5%:
r = 3.5%
rf = 1.5%
The volatility of the yen per dollar exchange rate is equal to the volatility of the dollar per
yen exchange rate. We convert the daily volatility into annual volatility:
s =
sh
h
=
0.00261712
1
365
= 0.00261712 ¥ 365 = 0.05
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SOA Sample Exam MFE/3F
Solutions
The values of d1 and d2 are:
d1 =
ln( x0 / K ) + (r - rf + 0.5s 2 )T
s T
=
ln(1) + (0.035 - 0.015 + 0.5 ¥ 0.052 )0.25
0.05 0.25
= 0.21250
d2 = d1 - s T = 0.21250 - 0.05 0.25 = 0.18750
We have:
N ( -d1 ) = N ( -0.21250) = 0.41586
N ( -d2 ) = N ( -0.18750) = 0.42563
The value of a put option on one yen is:
PEur (S , K , , r,T ,  )  Ke rT N ( d2 )  Se
rf T
N ( d1 )
1 0.035(0.25)
1 0.015(0.25)
 0.42563 
 0.41586
e
e
120
120
 0.0000634878

Since the option is for ¥120,000,000,000 the value of the put option in dollars is:
120,000,000,000 ¥ 0.0000634878 = 7,618,538
Solution 8
D
Chapter 8, Black-Scholes and Delta
Since the stock does not pay dividends, the price of the American call option is equal to
the price of the otherwise equivalent European call option.
Since the stock does not pay dividends, delta is equal to N (d1 ) :
DCall = e -d T N (d1 )
0.50  e00.25 N (d1 )
N (d1 )  0.50

d1  0.00
The formula for d1 can be solved for the risk-free rate of return:
d1 =
ln(S / K ) + ( r - d + 0.5s 2 )T
0.00 =
s T
ln(40 / 41.50) + ( r - 0 + 0.5 ¥ 0.302 )0.25
0.30 0.25
r = 0.102256
The value of d2 is:
d2 = d1 - s T = 0.00 - 0.30 0.25 = -0.15
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SOA Sample Exam MFE/3F
Solutions
The value of the call option is:
CEur (S , K , , r,T ,  )  Se  T N (d1 )  Ke rT N (d2 )
 40(0.5)  41.5e 0.1022560.25 N ( 0.15)
 20  40.453 1  N (0.15)
 20.453  40.453N (0.15)
 40.453N (0.15)  20.453
0.15 0.5 x 2
1
e
dx  20.453
 40.453 
2 

 16.138
0.15 0.5x 2

e
dx  20.453
Solution 9
B
Chapter 9, Market-Maker Profit
The market-maker’s profit is zero if the stock price movement is one standard deviation:
One standard deviation move = s St h
To answer the question, we must determine the value of s . We can determine the value
of N (d1 ) :
Call  e  T N (d1 )
0.61791  e 0.0(0.25) N (d1 )
0.61791  N (d1 )
Using the cumulative normal distribution calculator, this implies that:
d1 = 0.3
We can use the formula for d1 to solve for the value of s :
d1 
ln(S / K )  (r    0.5 2 )T
0.30 
 T
ln(50 / 50)  (0.10  0  0.5 2 )0.25
 0.25
0.15  0.025  0.125 2
0.125 2  0.15  0.025  0
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SOA Sample Exam MFE/3F
Solutions
We make use of the quadratic formula:
0.15 ± ( -0.15)2 - 4(0.125)(0.025)
2(0.125)
or
s = 0.20
s = 1.0
s =
The first of these volatility values, 0.20, seems more reasonable, and indeed, it produces
s St h = 0.20 ¥ 50 ¥
1
= 0.5234
365
If we use s = 1.0 , then we do not obtain one of the answer choices:
1
= 2.6171
365
s St h = 1.0 ¥ 50 ¥
Therefore, we conclude that s = 0.20 , and the stock price moves either up or down by
0.5234.
Solution 10
E
Chapter 14, Black-Scholes Framework
Statement (i) is true, because when the Black-Scholes framework applies, the natural log
of stock prices follows arithmetic Brownian motion.
Statement (ii) is true, because when the Black-Scholes framework applies, stock prices
are lognormally distributed:
(
ln S (t + h ) - ln S (t )  N (a - 0.5s 2 )h, s 2h
)
Since the stock prices are lognormally distributed, we have:
Var[ X (t + h ) - X (t )] = Var[ln S (t + h ) - ln S (t )] = s 2h
The quickest way to analyze statement (iii) is to convert the expression into its continuous
form:
n
Â
nÆ•
lim
j =1
2
ÈÎ X ( jT / n ) - X (( j - 1)T / n )˘˚ =
T
Ú [dX (t )]
2
0
The natural log of the stock prices follows arithmetic Brownian motion:
(
)
dX (t ) = d [ln S (t )] = a - d - 0.5s 2 dt + s dZ (t )
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SOA Sample Exam MFE/3F
Solutions
Substituting into the stochastic integral above, we have:
T
Ú [dX (t )]
2
0
T
=
Ú [d ln S(t )]
2
=
0
T
=
T
Ú ÈÎÍ(a - d - 0.5s
0
Ú
0
È(a - d - 0.5s 2 )dt + s dZ (t ) ˘
Î
˚
2
2
) dt 2 + 2(a - d - 0.5s 2 )dt ¥ s dZ (t ) + s 2 (dZ (t )) ˘
˚˙
2 2
We make use of the following multiplication rules to simplify the expression above:
dt 2 =0
and
dt ¥ dZ (t ) = 0
and
[dZ (t )]2 = 0
The stochastic integral can now be simplified:
T
Ú ÈÎÍ(a - d - 0.5s
0
T
=
Ú
0
2
) dt 2 + 2(a - d - 0.5s 2 )dt ¥ s dZ (t ) + s 2 (dZ (t )) ˘
˚˙
2 2
T
È 0 + 0 + s 2 dt ˘ = s 2 dt = s 2T
Î
˚
Ú
0
Therefore Statement (iii) is true. The correct answer is Choice E.
Alternate solution for Statement (iii)
Statement (iii) is true, because when the Black-Scholes framework applies, the natural
log of the stock prices follows arithmetic Brownian motion:
(
)
dX (t ) = d [ln S (t )] = a - d - 0.5s 2 dt + s dZ (t )
Therefore:
(
)
X (t ) = a - d - 0.5s 2 t + s Z (t )
Recall that we use Y (t ) to denote a random draw from a binomial approximation. For
small values of h, we have:
Z (t + h ) - Z (t ) = Y (t + h) h
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SOA Sample Exam MFE/3F
Solutions
We therefore can write:
n
Â ÈÎ X ( jT / n) - X (( j - 1)T / n)˘˚
n Æ•
lim
2
j =1
n
È a - d - 0.5s 2 jT / n + s Z ( jT / n ) - a - d - 0.5s 2 ( j - 1)T / n - s Z (( j - 1)T / n )˘
)
(
)
Â
ÍÎ (
˚˙
n Æ•
= lim
2
j =1
n
Â ÈÍÎ(a - d - 0.5s 2 ) T / n + s ÈÎZ ( jT / n) - Z (( j - 1)T / n)˘˚ ˘˙˚
n Æ•
= lim
2
j =1
n
(
)
È
Ê jT ˆ T ˘
2
= lim
Í a - d - 0.5s T / n + s Y Á
˙
Ë n ˜¯ n ˙˚
n Æ•
j =1 ÍÎ
Â
2
3
È
˘
2
2
Í
Ê jT ˆ s 2T ˙
2 ÊT ˆ
2 ÊT ˆ2
= lim
ÁË n ˜¯ + 2 a - d - 0.5s ÁË n ˜¯ s Y ÁË n ˜¯ + n ˙
Í a - d - 0.5s
n Æ•
j =1 Í
˙
Î
˚
3
˘
n È
n È
n È 2 ˘
2 Ê T ˆ2 ˘
s T
ÊT ˆ2
Ê jT ˆ ˙
Í
2
2
Í a - d - 0.5s
˙
a
d
s
s
Y
lim
2
0.5
lim
= lim
+
+
Í
˙
ÁË n ˜¯
ÁË n ˜¯
ÁË n ˜¯ ˙
Í
n
n Æ•
n Æ•
n Æ•
Í
˙
Í
˙˚
j =1 Î
j =1 Í
j =1 Î
˙
˚
Î
˚
n
Â
(
)
Â
(
)
(
)
Â
(
)
Â
We now consider each of the three terms separately.
The first of these terms is zero because:
È
È
2 Ê T ˆ2 ˘
2 Ê T ˆ2 ˘
Í a - d - 0.5s 2 Á ˜ ˙ = lim Í n ¥ a - d - 0.5s 2 Á ˜ ˙
Ë n ¯ ˙ nÆ• Í
Ë n¯ ˙
nÆ•
j =1 ÍÎ
˚
Î
˚
2
2
È1 ˘
2
2
= a - d - 0.5s 2 (T ) lim Í ˙ = a - d - 0.5s 2 (T ) ¥ 0 = 0
nÆ• Î n ˚
n
lim
Â
(
)
(
)
(
)
(
)
The second term is also zero. This is because both the maximum possible value and the
minimum possible value of the second term are zero. Consider the value that is obtained
if the binomial random variable Y ( jT / n ) is always equal to 1:
3/2
È
˘
ÊT ˆ
Í 2 a - d - 0.5s 2 Á ˜
s Y ( jT / n )˙
Ë n¯
n Æ•
˙˚
j =1 ÍÎ
n
lim
Â
(
)
3/2 ˘
È
ÊT ˆ
2
Í2 a - d - 0.5s Á ˜
s˙
= lim
Ë n¯
n Æ•
˙˚
j =1 ÍÎ
n
Â
(
)
3/2 ˘
È
1
ÊT ˆ
= lim Ín ¥ 2 a - d - 0.5s 2 Á ˜
s ˙ = 2 a - d - 0.5s 2 T lim
Ë
¯
n
n Æ• Í
n Æ• n
˙˚
Î
(
(
)
(
)
)
= 2 a - d - 0.5s 2 T ¥ 0 = 0
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SOA Sample Exam MFE/3F
Solutions
If the binomial variable Y ( jT / n ) is always -1 , then the second term is:
3/2
È
˘
ÊT ˆ
Í 2 a - d - 0.5s 2 Á ˜
s Y ( jT / n )˙
Ë n¯
nÆ•
˙˚
j =1 ÍÎ
n
lim
Â
(
)
3/2
È
˘
ÊT ˆ
Í2 a - d - 0.5s 2 Á ˜
s ¥ ( -1) ˙
Ë n¯
nÆ•
˙˚
j =1 ÍÎ
3/2 ˘
È
1
ÊT ˆ
s ˙ = -2 a - d - 0.5s 2 T lim
= - lim Ín ¥ 2 a - d - 0.5s 2 Á ˜
Ë
¯
n
nÆ• Í
nÆ• n
˙˚
Î
= lim
n
Â
(
)
(
)
(
(
)
)
= -2 a - d - 0.5s 2 T ¥ 0 = 0
Since the maximum possible value of the second term approaches zero, and the minimum
possible value of the second term approaches zero, we conclude that the second term
approaches zero as n becomes very large.
Finally, the third term is:
n
n
T
T
È 2T˘
= lim s 2
s
1] = lim s 2 n = lim s 2T = s 2T
[
Í
˙
n ˚ n Æ•
n j =1
n
n Æ•
n Æ•
n Æ•
j =1 Î
lim
Â
Â
Therefore, we have:
n
Â ÈÎ X ( jT / n) - X (( j - 1)T / n)˘˚
n Æ•
lim
2
= s 2T
j =1
The solution using the multiplication rules is much quicker than this alternate solution!
Solution 11
E
Chapter 14, Black-Scholes Framework
Statement (i) is true, because when the Black-Scholes framework applies, the stock prices
are lognormally distributed:
(
ln S (t + h ) - ln S (t )  N (a - d - 0.5s 2 )h, s 2h
)
If we are given S (t ) , this implies:
(
ln S (t + h )  N ln S (t ) + (a - d - 0.5s 2 )h, s 2h
)
Var ÈÎ ln( S (t + h ) S (t ) ˘˚ = s 2h
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12
SOA Sample Exam MFE/3F
Solutions
Statement (ii) is true, because when the Black-Scholes framework applies, the stock
dS (t )
= (a - d )dt + s dZ (t )
S (t )
The variance is:
È dS (t )
˘
Var Í
S (t ) ˙ = Var [(a - d )dt + s dZ (t )] = 0 + s 2Var [dZ (t )] = s 2dt
Î S (t )
˚
Statement (iii) is true, because when the Black-Scholes framework applies, the stock
dS (t ) = (a - d )S (t )dt + s S (t )dZ (t )
Therefore, we have:
Var ÈÎS (t + dt ) S (t ) ˘˚ = Var ÈÎS (t ) + dS (t ) S (t ) ˘˚ = 0 + Var ÈÎdS (t ) S (t ) ˘˚
= Var ÈÎ(a - d )S (t )dt + s S (t )dZ (t ) S (t ) ˘˚ = 0 + s 2 [S (t )] Var [dZ (t )]
2
= s 2 [S (t )] dt
2
Solution 12
B
Chapter 15, Sharpe Ratio
When the price follows geometric Brownian motion, the natural log of the price follows
arithmetic Brownian motion:
dS (t ) = a S (t )dt + s S (t )dZ (t )
¤
(
)
d [ln S (t )] = a - 0.5s 2 dt + s dZ (t )
Therefore:
dY (t )
= Adt + BdZ (t )
Y (t )
¤
(
)
d [ln Y (t )] = A - 0.5B 2 dt + BdZ (t )
The arithmetic Brownian motion provided in the question for d [ln Y (t )] allows us to solve
for B:
(
)
d [ln Y (t )] = m dt + 0.085dZ (t ) and d [ln Y (t )] = A - 0.5B 2 dt + BdZ (t )
A - 0.5B 2 = m
B = 0.085
Page
13
SOA Sample Exam MFE/3F
Solutions
Since X and Y have the same underlying source of risk, they must have the same Sharpe
ratio. This allows us to solve for A:
a X - r aY - r
=
sX
sY
0.07 - 0.04 A - 0.04
=
0.12
0.085
A = 0.06125
Solution 13
E
Chapter 15, Itô’s Lemma
The drift is the expected change in the asset price per unit of time.
(i) For the first equation, the partial derivatives are:
UZ  2
U ZZ  0
Ut  0
This results in:
1
dU (t ) = U Z dZ + U ZZ ( dZ )2 + Ut dt = 2dZ + 0 + 0 = 2dZ
2
Since there is no dt term, the drift is zero for dU.
(ii) For the second equation, the partial derivatives are:
VZ = 2Z
VZZ = 2
Vt = -1
This results in:
1
1
dV (t ) = VZ dZ + VZZ (dZ )2 + Vt dt = 2Z (t )dZ + ¥ 2(dZ )2 - 1dt
2
2
= 2Z (t )dZ + 1dt - 1dt = 2Z (t )dZ
Since there is no dt term, the drift is zero for dV.
(iii) We must find:
dW (t ) = d Èt2Z (t ) ˘ - 2tZ (t )dt
Î
˚
For the first term, we must use Itô’s Lemma. Let’s define:
X (t ) = t2Z (t )
The partial derivatives are:
X Z = t2
X ZZ = 0
Xt = 2tZ
Page
14
SOA Sample Exam MFE/3F
Solutions
This results in:
dX (t ) = X Z dZ +
1
1
X ZZ (dZ )2 + Xt dt = t 2dZ + ¥ 0 ¥ ( dZ )2 + 2tZdt
2
2
= t 2dZ (t ) + 2tZ (t )dt
Therefore:
dW (t ) = d Èt 2Z (t ) ˘ - 2tZ (t )dt = dX (t ) - 2tZ (t )dt = t2dZ (t ) + 2tZ (t )dt - 2tZ (t )dt
Î
˚
= t 2dZ (t )
Since there is no dt term, the drift is zero for dW.
Solution 14
0.0517 Chapter 19, Vasicek Model
The Vasicek model of short-term interest rates is:
dr = a(b - r )dt + s dZ
Therefore, we can determine the value of a:
dr = 0.6(b - r )dt + s dZ
ﬁ
a = 0.6
In the Vasicek model, the Sharpe ratio is constant:
f (r, t ) = f
Therefore, for any r, t, and T, we have:
f=
a (r, t,T ) - r
q( r , t , T )
Since the Sharpe ratio is constant:
a (0.04,0,2) - 0.04
q(0.04,0,2)
=
a (0.05,1,4) - 0.05
q(0.05,1,4)
We now make use of the following formula for q(r , t,T ) in the Vasicek model:
q(r , t,T ) = B(t,T )s (r ) = B(t ,T )s
Page
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SOA Sample Exam MFE/3F
Solutions
Substituting this expression for q(r, t,T ) into the preceding equality allows us to solve for
a (0.05,1,4) :
 (0.04,0,2)  0.04
q(0.04,0,2)

 (0.05,1,4)  0.05
q(0.05,1,4)
 (0.04,0,2)  0.04  (0.05,1,4)  0.05

B(0,2)
B(1,4)
0.04139761  0.04
1e
0.6(2  0)
0.6

0.00139761
1e

0.6(2  0)

 (0.05,1,4)  0.05
1  e 0.6(4 1)

0.6
 (0.05,1,4)  0.05
1  e 0.6(4 1)
0.00139761  (0.05,1,4)  0.05

0.6988
0.8347
 (0.05,1,4)  0.0516694
Solution 15
1.3264 Chapter 18, Black-Derman-Toy Model
In each column of rates, each rate is greater than the rate below it by a factor of:
e2s i h
Therefore, the missing rate in the third column is:
0.135
rdd = 0.135e -2s 2 1 = 0.135 ¥
= 0.106
0.172
The missing rates in the fourth column are:
0.168
rduu = 0.168e -2s 3 1 = 0.168 ∏
= 0.1359
0.110
0.168
rddd = 0.11e -2s 3 1 = 0.110 ∏
= 0.0890
0.110
The tree of short-term rates is:
16.80%
17.20%
12.60%
9.00%
13.59%
13.50%
9.30%
11.00%
10.60%
8.90%
Page
16
SOA Sample Exam MFE/3F
Solutions
The caplet pays off only if the interest rate at the end of the third year is greater than
10.5%. The payoff table is:
Time 0
Time 1
Time 2
Time 3
5.3938
0.0000
0.0000
0.0000
2.7238
0.0000
0.0000
0.4505
0.0000
0.0000
The payments have been converted to their equivalents payable at the end of 3 years. The
calculations for the tree above are shown below:
100 ¥ (0.1680 - 0.105)
= 5.3938
1.1680
100 ¥ (0.1359 - 0.105)
= 2.7238
1.1359
100 ¥ (0.1100 - 0.105)
= 0.4505
1.1100
We work recursively to calculate the value of the caplet. For example, if the interest rate
increases at the end of the first year and at the end of the second year, then its value at
the end of the second year is:
0.5 ¥ 5.3938 + 0.5 ¥ 2.7238
= 3.4632
1.1720
The completed tree is:
Time 0
Time 1
Time 2
Time 3
5.3938
3.4632
2.1588
1.3264
2.7238
1.3984
0.7329
0.4505
0.2036
0.0000
The value of the year-4 caplet is \$1.3264.
Page
17
SOA Sample Exam MFE/3F
Solutions
Solution 16
Chapter 15, Prepaid Forward Price of S a
B
We make use of the following expression for the prepaid forward price of the derivative:
FtP,T
 x ( r  )  0.5 x ( x 1) 2  (T t )

S (T )x   e r(T t ) S (t )x e 


We can substitute this expression for the prepaid forward price in the equation provided
in the question:
FtP,T S (T )x   S (t )x


e
r(T t )
2
x  x ( r  )  0.5x ( x 1)  (T t )
e
 S (t ) 
 r  x ( r  )  0.5 x ( x 1) 2  (T t )

e
 S (t )x
1
 r  x ( r   )  0.5x ( x  1) 2  (T  t )  0


r  x (r   )  0.5x ( x  1) 2  0
Putting in the risk-free interest rate of 4%, the volatility of 20% and the dividend yield of
0%, we have:
r  x (r   )  0.5x ( x  1) 2  0
0.04  x (0.04  0.00)  0.5x 2 (0.20)2  0.5x (0.20)2  0
0.02x 2  0.02x  0.04  0
x2  x  2  0
( x  1)( x  2)  0
x 1
or
x  2
We were told in the question that 1 is a solution. The other solution is -2.
Solution 17
A
Chapter 13, Estimating Volatility
Notice that the price increases by 25% 4 times and it decreases by 25% 4 times. This
means that the mean return is zero, and each of the deviations is the same size. This
simplifies the calculations below.
Since we have 9 months of data, we can calculate 8 monthly returns, and k = 8 . Each
monthly return is calculated as a continuously compounded rate:
Ê S ˆ
ri = ln Á i ˜
Ë Si -1 ¯
Page
18
SOA Sample Exam MFE/3F
Solutions
The next step is to calculate the average of the returns:
k
r =
Â ri
i =1
k
The returns and their average are shown in the third column below:
Month
Price
1
2
3
4
5
6
7
8
9
80
64
80
64
80
100
80
64
80
Ê S ˆ
ri = ln Á i ˜
Ë Si -1 ¯
(ri - r )2
–0.22314
0.22314
–0.22314
0.22314
0.22314
–0.22314
–0.22314
0.22314
0.049793
0.049793
0.049793
0.049793
0.049793
0.049793
0.049793
0.049793
r = 0.00000
Â (ri - r )2 = 0.398344
8
i =1
The fourth column shows the squared deviations and the sum of squares.
The estimate for the standard deviation of the monthly returns is:
k
sh =
Â (ri - r )2
i =1
s1 =
12
k -1
0.398344
= 0.23855
7
We adjust the monthly volatility to obtain the annual volatility. Since h =
1
:
12
1
= 0.23855 12 = 0.82636
h
s = sh
This problem isn’t very difficult if you are familiar with the statistical function of your
calculator.
On the TI-30X IIS, the steps are:
[2nd][STAT]
(Select 1-VAR)
[ENTER]
[DATA]
Ê 64 ˆ
[ENTER]
80 ˜¯
X1= ln Á
Ë
ØØ
(Hit the down arrow twice)
Page
19
SOA Sample Exam MFE/3F
Solutions
Ê 80 ˆ
[ENTER]
64 ˜¯
ØØ
Ê 64 ˆ
[ENTER]
Ë 80 ˜¯
ØØ
Ê 80 ˆ
[ENTER]
64 ˜¯
ØØ
Ê 100 ˆ
[ENTER]
80 ˜¯
ØØ
Ê 80 ˆ
[ENTER]
100 ˜¯
ØØ
Ê 64 ˆ
[ENTER]
Ë 80 ˜¯
ØØ
X2= ln Á
Ë
X3= ln Á
X4= ln Á
Ë
X5= ln Á
Ë
X6= ln Á
Ë
X7= ln Á
Ê 80 ˆ
[ENTER]
64 ˜¯
X8= ln Á
Ë
[STATVAR]
Æ Æ
¥
[ENTER]
(12)
(Arrow over to Sx)
The result is: 0.826363140
To exit the statistics mode:
[2nd] [EXITSTAT]
[ENTER]
On the BA II Plus calculator, the steps are:
[2nd][DATA]
[2nd][CLR WORK]
64/80
=
LN
[ENTER]
ØØ (Hit the down arrow twice)
80/64
=
LN
[ENTER]
ØØ
64/80
=
LN
[ENTER]
ØØ
80/64
=
LN
[ENTER]
ØØ
100/80 =
LN
[ENTER]
ØØ
80/100 =
LN
[ENTER]
ØØ
64/80
=
LN
[ENTER]
ØØ
80/64
=
LN
[ENTER]

¥
[2nd][STAT]
(12)
=
The result is: 0.82636314
To exit the statistics mode: [2nd][QUIT]
Page
20
SOA Sample Exam MFE/3F
Solutions
Solution 18
A
Chapter 10, Delta-Hedging Gap Call Options
The price of the gap call option is:
CGapCall = Se -d T N (d1 ) - K1 e -rT N (d2 )
CGapCall = SN ( d1 ) - 130 N (d2 )
The delta of the option is the derivative of the option price with respect to the stock’s
price:
DGapCall =
∂CGapCall
∂S
∂N ( d1 )
∂N (d2 )
- 130
∂S
∂S
∂N ( d1 ) ∂d1
∂N (d2 ) ∂d2
= N (d1 ) + S
- 130
∂d1 ∂S
∂d2 ∂S
= N (d1 ) + S
= N (d1 ) + SN '(d1 )
∂d1
∂d
- 130 N '(d2 ) 2
∂S
∂S
Note that:
N (d ) 
d

1
2
2
e 0.5x dx

N '(d ) 
1
2
e 0.5d
2
The derivatives of d1 and d2 with respect to the stock’s price are shown below:
d1 =
Ê Se -d T ˆ s 2
T
ln Á
˜+
2
Ë K 2e -rT ¯
s T
d2 = d1 - s T ﬁ
=
Ê e -d T ˆ s 2
T
ln (S ) + ln Á
˜+
2
Ë K 2e -rT ¯
s T
ﬁ
∂d1
1
1
=
=
∂S Ss T S
1
1
∂d2
=
=
∂S Ss T S
Let’s calculate the values of d1 and d2 that we’ll need for the formula for delta:
d1 =
Ê Se -d T ˆ s 2
T
ln Á
˜+
2
Ë K 2 e -rT ¯
s T
=
Ê 100e -0 ¥1 ˆ 12
ln Á
¥1
˜+
Ë 100e -0 ¥1 ¯ 2
1 1
= 0.5 ﬁ
N (d1 ) = 0.69146
d2 = d1 - s T = 0.5 - 1 1 = -0.5
Page
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SOA Sample Exam MFE/3F
Solutions
Now we can calculate the delta of the gap call option:
∂d1
∂d
- 130 N '(d2 ) 2
∂S
∂S
2
1 -0.5d1 1
1 -0.5d22 1
= 0.69146 + S
e
- 130
e
S
S
2p
2p
1 -0.5 ¥ 0.52
1 -0.5( -0.5)2 1
= 0.69146 +
e
- 130
e
100
2p
2p
DGapCall = N (d1 ) + SN '(d1 )
= 0.69146 +
1
2p
2
e -0.5 ¥ 0.5 (1 - 1.3)
= 0.58584
Since the market-maker sells 1,000 of the gap call options, the market-maker multiplies
the delta of one call option by 1,000 to determine the number of shares that must be
purchased in order to delta-hedge the position:
1,000 ¥ 0.58584 = 585.84
Solution 19
C
Chapter 11, Forward Start Option
In one year, the value of the call option will be:
CEur (S1 ) = S1e -d T N (d1 ) - Ke -rT N (d2 )
= S1 N (d1 ) - S1e -0.08 N (d2 )
In one year, the values of d1 and d2 will be:
d1 =
ln( S / K ) + (r - d + 0.5s 2 )T
s T
=
ln(S1 / S1 ) + (0.08 - 0.00 + 0.5 ¥ 0.302 )(1)
0.30 1
= 0.41667
d2 = d1 - s T = 0.41667 - 0.30 1 = 0.11667
From the normal distribution table:
N ( d1 ) = N (0.41667) ª 0.66154
N ( d2 ) = N (0.11667) ª 0.54644
In one year, the value of the call option can be expressed in terms of the stock price at
that time:
CEur (S1 ) = S1 N (d1 ) - S1 e -0.08 N (d2 )
= S1 ¥ 0.66154 - S1 e -0.08 ¥ 0.54644
= 0.15711 ¥ S1
In one year, the call option will be worth 0.15711 shares of stock.
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SOA Sample Exam MFE/3F
Solutions
The prepaid forward price of one share of stock is:
F0,PT (S ) = e -rT F0,T (S )
P
F0,1
(S ) = e -0.08 F0,1 ( S )
P
F0,1
(S ) = e -0.08 ¥ 100
P
F0,1
(S ) = 92.31163
The value today of 0.15711 shares of stock in one year is:
0.15711 ¥ 92.31163 = 14.5033
Solution 20
D
Chapter 8, Portfolio Delta and Elasticity
The elasticity of portfolio A is the weighted average of the elasticities of its components:
WPort =
n
Â wi Wi
i =1
2 ¥ 4.45
1.90
5=
WCall +
WPut
2 ¥ 4.45 + 1.90
2 ¥ 4.45 + 1.90
54 = 8.90WCall + 1.90W Put
We can express the delta of an option in terms of the elasticity of the option:
W=
SD
V
ﬁ
D=W
V
S
The delta of portfolio B is the sum of the deltas of its components:
D Port =
n
Â qi Di
i =1
3.4 = 2DCall - 3D Put
4.45
1.90
- 3W Put
45
45
153 = 8.90WCall - 5.70W Put
3.4 = 2WCall
We now have system of 2 equations with 2 unknowns:
54 = 8.90WCall + 1.90W Put
153 = 8.90WCall - 5.70WPut
Let’s subtract the second equation from the first equation:
-99 = 7.60WPut
W Put = -13.0263
Page
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SOA Sample Exam MFE/3F
Solutions
Solution 21
C
Chapter 19, Cox-Ingersoll-Ross Model
We begin with the Sharpe ratio and parameterize it for the CIR model:
f (r, t ) =
f
r
=
s
fr =
a (r, t,T ) - r
q( r , t , T )
a (r, t,T ) - r
B(t ,T )s ( r )
a (r , t,T ) - r
B(t,T )
We use the value of a (0.05,7,9) provided in the question:
f ¥ 0.05 =
a (0.05,7,9) - 0.05
B(7,9)
0.06 - 0.05
f ¥ 0.05 =
B(7,9)
0.01 = B(7,9)f (0.05)
B(7,9)f = 0.20
Making use of the fact that B(7,9) = B(11,13) , we have:
f ¥ 0.04 =
a (0.04,11,13) - 0.04
B(11,13)
a (0.04,11,13) = 0.04 + B(11,13)f (0.04)
= 0.04 + 0.20(0.04)
= 0.048
Solution 22
D
Chapter 19, Risk-Neutral Process & Sharpe Ratio
The realistic process for the short rate follows:
dr = a(r )dt + s (r )dZ
where:
a(r ) = 0.09 - 0.5r
&
s (r ) = 0.3
The risk-neutral process follows:
dr = [a( r ) + s (r )f (r, t )] dt + s (r )dZ
We can use the coefficient of the first term of the risk-neutral process to solve for the
Sharpe ratio, f (r , t ) :
0.15 - 0.5r = a( r ) + s (r )f (r, t )
0.15 - 0.5r = 0.09 - 0.5r + 0.3f (r, t )
f (r, t ) = 0.20
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SOA Sample Exam MFE/3F
Solutions
The derivative, like all interest-rate dependent assets in this model, must have a Sharpe
ratio of 0.20. Let’s rearrange the differential equation for g, so that we can more easily
observe its Sharpe ratio:
dg m
= dt - 0.4dZ
g
g
m
ﬁ
-r
g
f ( r, t ) =
0.4
Since the Sharpe ratio is 0.20:
m
-r
g
0.20 =
0.4
0.08 =
m
-r
g
m = (r + 0.08) g
Solution 23
C
Chapter 16, Sharpe Ratio
The stock and the call option must have the same Sharpe ratio:
0.10 - 0.04
s
=
g - 0.04
sC
The cost of the shares required to delta-hedge the call option is the number of shares
required, D = VS , times the cost of each share, S. Therefore, based on statement (iv) in
the question, we have:
VS ¥ S = 9
We can find the volatility of the call option in terms of the volatility of the underlying
stock:
sC =
SVS
9
s = s = 1.5s
V
6
We can now solve for g :
0.10 - 0.04
s
=
g - 0.04
1.5s
ﬁ
g = 0.13
Page
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SOA Sample Exam MFE/3F
Solutions
Solution 24
E
Chapter 14, Ornstein-Uhlenbeck Process
Let’s find the differential form of each choice.
Choice A does not contain a random variable, so its differential form is easy to find:
dX (t ) = È - l X (0)e - lt + al e - lt ˘ dt
Î
˚
Choice B is easily recognized as an arithmetic Brownian motion:
dX (t ) = a dt + s dZ (t )
Choice C is easily recognized as a geometric Brownian motion:
dX (t ) = a X (t )dt + s X (t )dZ (t )
Choice D is:
dX (t ) = al e lt dt + s e lt dZ (t )
Choice E contains 3 terms. The first two terms do not contain random variables, so their
differentials are easy to find. The third term will be more difficult:
dX (t ) = - l X (0)e
- lt
dt + la e
- lt
Èt
˘
dt + d Í s e - l (t - s )dZ ( s) ˙
Í
˙
Î0
˚
Ú
The third term has a function of t in the integral. We can pull the t-dependent portion out
of the integral, so that we are finding the differential of a product. We then use the
following version of the product rule to find the differential:
d [U (t )V (t )] = dU (t )V (t ) + U (t )dV (t )
The differential of the third term is:
t
È t
˘
È
˘
- l (t - s )
- lt
Í
˙
Í
d s e
dZ ( s) = s d e
e l sdZ ( s) ˙
Í
˙
Í
˙
0
Î 0
˚
Î
˚
Ú
Ú
t
Èt
˘
- lt ˘
ls
- lt Í l s
È
e dZ ( s) + s e d e dZ ( s) ˙
=sd e
Î
˚
Í
˙
0
Î0
˚
t
Ê
ˆ
= - ls e - lt Á e l sdZ ( s)˜ dt + s e - lt e lt dZ (t )
ÁË
˜¯
0
Ú
Ú
Ú
Êt
ˆ
= - ls Á e - l (t - s )dZ ( s )˜ dt + s dZ (t )
ÁË
˜¯
0
Ú
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SOA Sample Exam MFE/3F
Solutions
Putting the three differentials together, we have:
dX (t ) = - l X (0)e
- lt
dt + la e
- lt
Êt
ˆ
dt - ls Á e - l (t - s )dZ ( s)˜ dt + s dZ (t )
ÁË
˜¯
0
Ú
È
Êt
ˆ˘
= - l Í X (0)e - lt - a e - lt + s Á e - l (t - s )dZ ( s)˜ ˙ dt + s dZ (t )
Í
ÁË
˜¯ ˙
0
Î
˚
t
È
˘
Ê
ˆ
= - l Í X (0)e - lt + a - a e - lt + s Á e - l (t - s )dZ ( s)˜ - a ˙ dt + s dZ (t )
Í
˙
ÁË
˜¯
0
Î
˚
È
˘
Êt
ˆ
- lt
- l (t - s )
- lt
Í
dZ ( s)˜ - a ˙ dt + s dZ (t )
= - l X (0)e
+ a (1 - e ) + s Á e
Í
˙
ÁË
˜¯
0
Î
˚
= - l [ X (t ) - a ]] dt + s dZ (t )
Ú
Ú
Ú
= l [a - X (t )]] dt + s dZ (t )
Solution 25
B
Chapter 11, Chooser Options
The price of the chooser option can be expressed in terms of a call option and put option:
(
Price of Chooser Option = CEur (S0 , K ,T ) + e -d (T -t1 ) PEur S0 , Ke -( r -d )(T -t1 ) , t1
)
The risk-free interest rate and the dividend yield are both zero, so the put option has the
same strike price as the call option:
(
Price of Chooser Option = CEur (S0 , K ,T ) + e -d (T -t1 ) PEur S0 , Ke -( r -d )(T -t1 ) , t1
)
Price of Chooser Option = CEur (95,100,3) + PEur (95,100,1)
We can use put-call parity to find the value of the 1-year call option:
CEur (95,100,1) + K = S0 + PEur (95,100,1)
4 + 100 = 95 + PEur (95,100,1)
PEur (95,100,1) = 9
We can now solve for the price of the 3-year call option:
Price of Chooser Option = CEur (95,100,3) + PEur (95,100,1)
20 = CEur (95,100,3) + 9
CEur (95,100,3) = 11
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Solution 26
D
Chapter 2, Bounds on Option Prices
Let’s begin by noting the prepaid forward price of the stock and the present value of the
strike price:
FtP,T (S ) = e -d (T -t )St = e -0 ¥ (0.5)St = St
Ke - r(T -t ) = 100e -0.10(0.5) = 95.12
Since the stock does not pay dividends, the American call option has the same price as the
European call option, so both call options correspond to the same graph. The value of a
call increases with the stock price, so the call options must correspond to either Graph I or
Graph II. For call options we have:
Max È0, FtP,T (S ) - Ke -r(T -t ) ˘ £ CEur (St , K ,T - t ) £ C Amer (St , K ,T - t ) £ St
Î
˚
Max [0, St - 95.12] £ CEur (St , K ,T - t ) £ C Amer (St , K ,T - t ) £ St
The left portion of the inequality above describes the lower boundary of Graph II. The
right portion of the inequality above describes the upper boundary of Graph II. Therefore,
the call options correspond to Graph II, which narrows the answer choices to (D) and (E).
For the put options, we have:
Max È0, Ke - r(T -t ) - FtP,T (S ) ˘ £ PEur (St , K ,T - t ) £ PAmer (St , K ,T - t ) £ K
Î
˚
Max [0,95.12 - St ] £ PEur (St , K ,T - t ) £ PAmer (St , K ,T - t ) £ 100
For the European put option, the left portion of the inequality above describes the lower
boundary of Graph IV. From the inequality above, we can see that it is possible for the
European put option to have a tighter upper bound than the American put option. In fact,
a European put option cannot have a price greater than the present value of the strike
price. Therefore, the upper boundary is \$95.12, and we can conclude that Graph IV
corresponds to the European put option, which still leaves answer choices (D) and (E).
The American put option could have a price as high as the strike price of \$100 if the stock
price falls to zero. This is described by the upper boundary of Graph III. From the
inequality above, we can see that it is possible for the American put option to have a
tighter lower bound than the European put option. In fact, since the American put option
can be exercised at any time, the lower boundary is the maximum of zero and the exercise
value, as shown in Graph III. Therefore, Graph III corresponds to the American put
option, and choice (D) is correct.
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Solutions
Solution 27
A
Chapter 3, Replication
The end-of-year payoffs of the call and put options in each scenario are shown in the table
below. The rightmost column is the payoff resulting from buying the put option and
selling the call option.
End of Year
Price of
Stock X
End of Year
Price of
Stock Y
CX
PY
PY - CX
Payoff
Payoff
Payoff
Outcome 1
\$200
\$0
105
95
–10
Outcome 2
\$50
\$0
0
95
95
Outcome 3
\$0
\$300
0
0
0
We need to determine the cost of replicating the payoffs in the rightmost column above.
We can replicate those payoffs by determining the proper amount of Stock X, Stock Y, and
the risk-free asset to purchase.
Let’s define the following variables:
X = Number of shares of Stock X to purchase
Y = Number of shares of Stock Y to purchase
Z = Amount to lend at the risk-free rate
We have 3 equations and 3 unknown variables:
Outcome 1:
200 X +
Outcome 2:
50 X +
0Y + Ze0.10 = -10
0Y + Ze0.10 = 95
0 X + 300Y + Ze0.10 = 0
Outcome 3:
Subtracting the second equation from the first equation allows us to find X:
150 X = -105
X = -0.7
We can put this value of X into the first equation to find the value of Z:
200( -0.7) + Ze0.10 = -10
Ze0.10 = 130
Z = 117.6289
We can use this value of Z in the third equation to find the value of Y:
300Y + 117.629e0.10 = 0
Y = -0.4333
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The cost now of replicating the payoffs resulting from buying the put and selling the call
is equal to the cost of establishing a position consisting of X shares of Stock X, Y shares of
Stock Y, and Z invested at the risk-free rate:
100 X + 100Y + Z = 100 ¥ ( -0.7) + 100 ¥ ( -0.4333) + 117.6289 = 4.30
Solution 28
A
Chapter 11, All-or-Nothing Options
For the square of the final stock price to be greater than 100, the final stock price must be
greater than 10:
[S(1)]2 > 100
¤
S (1) > 10
Therefore, the option described in the question is 100 cash-or-nothing call options that
have a strike price of 10. The current value of the option is:
100 ¥ CashCall(S , K ,T ) = 100 ¥ e - rT N (d2 ) = 100 ¥ e -0.02 N (d2 )
To find the delta of the option, we must find the derivative of the price with respect to the
stock price:
(
∂ 100 ¥ e -0.02 N ( d2 )
∂S
) = 100e
-0.02
∂ ( N (d2 ))
∂S
= 100e -0.02 N '( d2 )
∂d2
∂S
The derivative of d2 with respect to the stock price is:
Ê Ê Se -d T ˆ s 2
ˆ
T
Á ln Á - rT ˜ +
˜
2
¯
Á Ë Ke
˜
∂Á
-s T ˜
s T
Á
˜ Ke -rT
e -d T
¥
Á
˜
Ë
¯ Se -d T Ke -rT
∂d2 ∂ d1 - s T
1
=
=
=
=
∂S
∂S
∂S
Ss T
s T
1
=
= 0.5
10 ¥ 0.20 ¥ 1
(
)
The current value of d2 is:
d2 =
Ê Se -d T ˆ s 2
ln Á
T
˜2
Ë Ke -rT ¯
s T
Ê 10 ˆ 0.22
ln Á
2
Ë 10e -0.02 ¯˜
=
=0
0.20
The density function for the standard normal random variable is:
N '( x ) =
1
2p
e -0.5x
2
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SOA Sample Exam MFE/3F
Solutions
We can now calculate the delta of the option:
100e -0.02 N '( d2 )
∂d2
1 -0.5d22
1
e
= 100e -0.02
¥ 0.5 = 100e -0.02
¥ e0 ¥ 0.5
∂S
2p
2p
= 19.552
Therefore, 19.552 shares must be purchased to delta-hedge the option.
Solution 29
D
Chapter 18, Black-Derman-Toy Model
Each node is e2s i h times the one below it. This means that 6% is e4s 2 times as large as
2%:
0.06 = 0.02e4s 2
Ê 0.06 ˆ 1
¥ = 0.27465
Ë 0.02 ˜¯ 4
s 2 = ln Á
The interest rate for the node above 2% is:
0.02e2s 2 = 0.02e2 ¥ 0.27465 = 0.034641
The tree of prices for the 3-year bond is:
0.9434
?
0.9095
0.9665
0.9451
0.9804
The calculations to obtain these prices are:
P (2,3, ruu ) =
1
= 0.9434
1.06
P (2,3, rud ) = P (2,3, rdu ) =
1
= 0.9665
1.034641
1
= 0.9804
1.02
0.5(0.9434 + 0.9665)
= 0.9095
P (1,3, ru ) =
1.05
0.5(0.9665 + 0.9804)
= 0.9451
P (1,3, rd ) =
1.03
P (2,3, rdd ) =
We cannot calculate the current price of the bond, P(0,3) because we do not know the
value of r0 , but we do not need P(0,3) to answer this question.
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The formula for the one-year yield volatility for a T-year zero-coupon bond is:
È P (1,T , r )-1 /(T -1) - 1 ˘
u
Yield volatility = 0.5 ¥ ln Í
˙
ÍÎ P (1,T , rd )-1 /(T -1) - 1 ˙˚
The yield volatility of the 3-year bond is:
È 0.9095-1 /(3 -1) - 1 ˘
Yield volatility = 0.5 ¥ ln Í
˙ = 0.26435
ÍÎ 0.9451-1 /(3 -1) - 1 ˙˚
Solution 30
A
Chapter 18, Black-Derman-Toy Model
The value of r0 is the same as the yield on the 1-year zero-coupon bond, and the yield
volatility of the 2-year bond is s1 :
1
= 0.9434
1 + r0
ﬁ
r0 = 6%
s1 = 10%
The first part of the interest rate tree is:
ru = R1e2s1
r0
rd e0.20
ﬁ
6%
rd = R1
rd
We need to determine the value of rd . The tree must correctly price the 2-year zerocoupon bond, so:
P (0,2) =
Ê
1
1
1 ˆ
(0.5) Á
+
0.20 1 + r ˜
1.06
Ë 1 + rd e
d¯
Ê
1
1 ˆ
0.8850 = 0.9434(0.5) Á
+
0.20 1 + r ˜
d¯
Ë 1 + rd e
The quickest way to answer this question is to use trial and error.
Let’s try Choice C first:


1
1

 0.8754
0.9434(0.5) 
 1  (0.07)e0.20 1  0.07 


Since the 0.8754 is less than 0.8850, let’s try a lower short rate.
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SOA Sample Exam MFE/3F
Solutions
Let’s try Choice A:


1
1
0.9434(0.5) 

 0.8850
 1  (0.0594)e0.20 1  0.0594 


Choice A is the correct answer.
Solution 31
B
Chapter 8, Delta
The delta of a bull spread is the same regardless of whether it is constructed of calls or
puts. Let’s assume that the bull spread consists of calls.
Since the bull spread consists of purchasing the lower-strike call and selling the higherstrike call, the delta of the bull spread is:
D50 - D60
We can find the values of delta using:
DCall = e -d T N (d1 )
and
d1 =
ln(S / K ) + (r - d + 0.5s 2 )T
s T
With T = 0.25 and K = 50 , we have:
d1 =
(
)
ln(50 / 50) + 0.05 + 0.5(0.20)2 (0.25)
0.20 0.25
N (0.17500) = 0.56946
= 0.17500
D50 = e -0 ¥ 0.25 (0.56946) = 0.56946
With T = 0.25 and K = 60 , we have:
d1 =
(
)
ln(50 / 60) + 0.05 + 0.5(0.20)2 (0.25)
0.20 0.25
N ( -1.64822) = 0.04965
= -1.64822
D60 = e -0 ¥ 0.25 (0.04965) = 0.04965
Therefore, the delta of the bull spread is initially:
D50 - D60 = 0.56946 - 0.04965 = 0.51981
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With T = 2 /12 and K = 50 , we have:
d1 =
(
ln(50 / 50) + 0.05 + 0.5(0.20)2
0.20
2
12
) 122 = 0.14289
N (0.14289) = 0.55681
D50 = e -0 ¥ 2 / 12 (0.55681) = 0.55681
With T = 2 /12 and K = 60 , we have:
d1 =
(
ln(50 / 60) + 0.05 + 0.5(0.20)2
2
12
N ( -2.09009) = 0.01830
) 122 = -2.09009
0.20
D60 = e -0 ¥ 2 / 12 (0.01830) = 0.01830
Therefore, the delta of the bull spread after 1 month is:
D50 - D60 = 0.55681 - 0.01830 = 0.53851
The change in the delta of the bull spread is:
0.53851 - 0.51981 = 0.01870
Solution 32
E
Chapter 14, Geometric Brownian Motion
The instantaneous return on the mutual fund is the weighted average of the return on the
stock and the return on the risk-free asset:
dW (t )
dS (t )
=j
+ (1 - j )rdt
W (t )
S (t )
= j [a dt + s dZ (t )] + (1 - j )rdt
= [ja + (1 - j )r ] dt + js dZ (t )
The expression above does not match Choice (A), because the final term in Choice (A) does
not include the factor j . Therefore, we can rule out Choice (A).
As written above, we see that W (t ) is a geometric Brownian motion. Therefore, the price
can be expressed as:
W (t )
Èja + (1 -j )r - 0.5j 2s 2 ˘t +js Z (t )
˚
= W (0)e Î
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Solutions
Although the expression above is close to Choices (B) and (C), it is slightly different from
both choices. Therefore, we turn to Choices (D) and (E).
Since Choices (D) and (E) contain [S (t ) / S (0)]j , let’s find the value of this expression:
2
S (t ) = S (0)e(a - 0.5s )t +s Z (t )
j
È S (t ) ˘
j (a - 0.5s 2 )t +js Z (t )
Í S (0) ˙ = e
Î
˚
j
È S (t ) ˘
(ja - 0.5js 2 )t +js Z (t )
e
=
Í S (0) ˙
Î
˚
We can now describe the process of the mutual fund with:
Èja + (1 -j )r - 0.5j 2s 2 ˘t +js Z (t )
˚
W (t ) = W (0)e Î
Èja - 0.5js 2 ˘t +js Z (t ) È(1 -j )r - 0.5j 2s 2 + 0.5js 2 ˘t
˚
˚
eÎ
= W (0)e Î
j
2 2
2
È S (t ) ˘ ÈÎ(1 -j )r - 0.5j s + 0.5js ˘˚t
= W (0) Í
e
˙
Î S (0) ˚
j
2
È S (t ) ˘ ÈÎ(1 -j )r + 0.5js ( -j +1)˘˚t
e
= W (0) Í
˙
Î S (0) ˚
j
2
È S (t ) ˘ (1 -j ) ÈÎr + 0.5js ˘˚t
e
= W (0) Í
˙
Î S (0) ˚
Solution 33
C
Chapter 11, Forward Start Options
The values of d1 and d2 for a 3-month put option with a strike price that is 90% of the
current stock price are:
d1 
ln( S / 0.90S )  (r    0.5 2 )T
 T

 ln(0.90)  (0.08  0.00  0.5  0.32 )(0.25)
0.3 0.25
 0.91074
d2  d1   T  0.9107  0.3 0.25  0.76074
From the cumulative normal distribution calculator:
N ( d1 )  N ( 0.91074)  0.18122
N ( d2 )  N ( 0.76074)  0.22341
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Solutions
The cost of a 3-month put option that has a strike price of 90% of the current stock price is
a function of the then-current stock price:
PEur (S ,0.90S ,0.25) = 0.90Se -rT N ( -d2 ) - Se -d T N ( -d1 )
= 0.90Se -0.25 ¥ 0.08 N ( -d2 ) - Se -0.00 N ( -d1 )
= S È0.90e -0.02 (0.22341) - 0.18122 ˘
Î
˚
= 0.01587S
Therefore, each of the 3-month put options can be purchased with 0.01587 shares of stock.
The cost of purchasing all 4 of the options now is equal to the cost of acquiring
4 ¥ 0.01587 = 0.06347 shares of stock. Since the current stock price is \$45, this cost is:
0.06347 ¥ 45 = 2.8562
Solution 34
A
Chapter 14, Multiplication Rules
As n goes to infinity, we can replace the summation with an integral:
n
lim
n Æ•
Â
j =1
3
T
{Z[ jh] - Z[( j - 1)h]} = [dZ (t )]3
Ú
0
We can now use the multiplication rules to evaluate the integral:
T
Ú [dZ (t )]
3
0
T
Ú
2
= [dZ (t )] ¥ dZ (t ) =
0
T
T
0
0
Ú dt ¥ dZ (t ) = Ú 0 = 0
The expected value and variance of zero are both zero:
N (0,0)
Solution 35
B
Chapter 14, Itô’s Lemma
Since X  R2 , the partial derivatives of X (t ) are:
X R = 2R
X RR = 2
Xt = 0
We can use Itô’s Lemma to find an expression for the differential of X (t ) :
dX (t ) = X R dR(t ) +
1
X RR [ dR(t )]2 + X t dt
2
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SOA Sample Exam MFE/3F
Solutions
Let’s find dR(t ) . The first two terms of R(t ) do not contain random variables, so their
differentials are easy to find. The third term will be more difficult:
t
È
˘
dR(t ) = - R(0)e -t dt + 0.05e -t dt + d Í 0.1 e s -t R( s)dZ ( s ) ˙
0
Î
˚
Ú
The third term has a function of t in the integral. We can pull the t-dependent portion out
of the integral, so that we are finding the differential of a product. We then use the
following version of the product rule to find the differential:
d [U (t )V (t )] = dU (t )V (t ) + U (t )dV (t )
The differential of the third term is:
t
t
È
˘
È
˘
d Í0.1 e s -t R( s )dZ ( s) ˙ = d Í 0.1e -t es R( s)dZ ( s) ˙
0
0
Î
˚
Î
˚
Ú
Ú
= -0.1e -t dt
= -0.1e -t dt
t s
Ú0 e
t s
e
0
Ú
R( s)dZ ( s ) + 0.1e -t et R(t )dZ (t )
R( s)dZ ( s ) + 0.1 R(t )dZ (t )
Putting all three terms together, we have:
dR(t )   R(0)e t dt  0.05e t dt  0.1e t dt
t s
0 e
R( s)dZ ( s)  0.1 R(t )dZ (t )
t


   R(0)e t  0.05e t  0.1e t e s R( s)dZ ( s)  dt  0.1 R(t )dZ (t )
0


t


   R(0)e t  0.05  0.05e t  0.1 e s t R( s)dZ ( s)  dt  0.05dt  0.1 R(t )dZ (t )
0




t


   R(0)e t  0.05(1  e t )  0.1 e s t R( s)dZ ( s)  dt  0.05dt  0.1 R(t )dZ (t )
0



  R(t )dt  0.05dt  0.1 R(t )dZ (t )
Making use of the multiplication rules, we see that:
[dR(t )]2 = 0.01R(t )dt
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Solutions
We can now find the differential of X (t ) . To simplify the notation, we use R for R(t ) and
X for X (t ) below:
1
X RR (dR )2 + X t dt
2
1
= 2RdR + ¥ 2(dR )2 + 0dt
2
dX = X R dR +
= 2RdR + (dR )2
= 2R[ - Rdt + 0.05dt + 0.1 RdZ ] + 0.01Rdt
= -2R2dt + 0.10 Rdt + 0.2R3 / 2dZ + 0.01Rdt
= È0.11R - 2R2 ˘ dt + 0.2R3 / 2dZ
Î
˚
We are given that X is equal to R2 :
X = R2
ﬁ
X =R
Substituting for R, we have:
dX = È0.11R - 2R2 ˘ dt + 0.2R3 / 2dZ
Î
˚
= ÈÎ0.11 X - 2 X ˘˚ dt + 0.2 X 3 / 4 dZ
Solution 36
E
Chapter 16, Black-Scholes Equation
To simplify the notation, let's set the exponent in the price of the derivative security equal
to a:
2
V  S k /   S a
The partial derivatives are shown below:
VS  aS a 1
VSS  a(a  1)S a  2
Vt  0
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Solutions
The Black-Scholes Equation gives us:
0.5s 2S 2VSS + (r - d )SVS + Vt + D(t ) = rV
0.5s 2S 2a(a - 1)S a - 2 + (0.04 - 0)SaS a -1 + 0 + 0 = 0.04S a
0.5s 2a( a - 1) + (0.04 - 0)a = 0.04
0.5s 2a( a - 1) + 0.04a - 0.04 = 0
0.5s 2a( a - 1) + 0.04( a - 1) = 0
(a - 1)(0.5s 2a + 0.04) = 0
a - 1 = 0 or 0.5s 2a + 0.04 = 0
0.08
a = 1 or a = -
s2
The question specifies that k is positive, so a must be negative, and therefore a cannot be
equal to 1:
a
k

2
0.08

2
0.08
2
k  0.08
Solution 37
D
Chapter 14, Brownian Motion
We make use of the following equivalency:
dS (t )
= (a - d )dt + s dZ (t )
S (t )
¤
dS (t )
= 0.03dt + 0.2dZ (t )
S (t )
¤
(
È S (t + h ) ˘
2
2
ln Í
˙  N (a - d - 0.5s )h, s h
S
(
t
)
Î
˚
È S (t + h ) ˘
ln Í
˙  N (0.01h, 0.04h )
Î S (t ) ˚
)
To avoid a cluttered appearance, we use St to represent S (t ) in the solution below.
We can rewrite the expression for G , so that it is the product of independent random
variables:
1/3
G  S1  S2  S3 
1/3

S 
S S 
S S S 
  S0 1   S0 1 2   S0 1 2 3  
S0  
S0 S 1  
S0 S 1 S2  

1/3

S3 S 2 S 
 S03  1  2  3 
S03 S12 S2 

S
 S0   1
 S0
2
1
  S2  3  S3  3

 

 S2 
  S1 
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Solutions
Taking the natural log, we have a sum of independent random variables:
S  2 S  1 S 
ln G  ln  S0   ln  1   ln  2   ln  3 
 S0  3  S1  3  S2 
The variance is:
2
2
2
1 
Var  ln G   0  0.04    (0.04)    (0.04)  0.06222
3
3
Solution 38
B
Chapter 19, Delta-Gamma Approximation for Bonds
In the model described in the question, we have:
P (r, t,T ) = A(t ,T )e - B(t ,T )r
Pr (r , t ,T ) = - B(t ,T ) A(t,T )e - B(t ,T )r = - B(t ,T )P (r , t ,T )
Prr (r, t,T ) = [ B(t,T )] A(t,T )e - B(t ,T )r = [ B(t,T )] P (r, t,T )
2
2
The delta-gamma-theta approximation is:
P [r(t + h ), t + h,T ] - P [r(t ), t,T ] ª [r(t + h ) - r(t )] Pr + 0.5 [r(t + h) - r(t )] Prr + Pt h
2
This question asks us to use only the delta and gamma portions of the approximation, so
we remove the theta term, Pt h , from the expression above:
P [r(t + h), t + h,T ] - P [r(t ), t,T ] ª [r(t + h) - r(t )] Pr + 0.5 [r(t + h) - r(t )] Prr
2
We replace r(t ) with 0.05, and we replace r(t + h) with 0.03.
The delta-gamma approximation is therefore:
PEst (0.03,0,3) - P (0.05,0,3) ª [ -0.02] Pr (0.05,0,3) + 0.5 [ -0.02] Prr (0.05,0,3)
2
PEst (0.03,0,3) ª P (0.05,0,3) + [ -0.02] Pr (0.05,0,3) + 0.5 [ -0.02] Prr (0.05,0,3)
2
PEst (0.03,0,3) ª P (0.05,0,3)
- [ -0.02] B(0,3)P (0.05,0,3) + 0.5 [ -0.02]
2
[ B(0,3)]2 P (0.05,0,3)
PEst (0.03,0,3) ª P (0.05,0,3) È1 + 0.02 ¥ 2 + 0.5 ¥ ( -0.02)2 ¥ 22 ˘
Î
˚
PEst (0.03,0,3) ª P (0.05,0,3) ¥ 1.0408
Therefore, we have:
PEst (0.03,0,3) P (0.05,0,3) ¥ 1.0408
ª
= 1.0408
P (0.05,0,3)
P (0.05,0,3)
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Solutions
Solution 39
B
Chapters 3 & 18, Risk-Neutral Probability
Let’s use P (0,2) to denote the price of a 2-year zero-coupon bond that matures for \$1.
We can make use of put-call parity:
C (108) + 108 ¥ P (0,2) = S0 + P (108)
P (108) - C (108) = 108 ¥ P (0,2) - 100
We can use the stock prices to determine the risk-neutral probability that the up state of
the world occurs:
p* 
e( r  )h  d (1  r0 )h  d (1.05)1  0.95 0.10 2




ud
ud
1.10  0.95
0.15 3
1
at time 1, and
1.06
1
if the down state occurs, then the zero-coupon bond will have a value of
at time 1.
1.04
The time 0 value is found using the risk-neutral probabilities and the risk-free rate at
time 0:
If the up state occurs, then the zero-coupon bond will have a value of
P (0,2) =
1 È2
1
1
1 ˘
¥
+ ¥
= 0.90423
Í
1.05 Î 3 1.06 3 1.04 ˙˚
We can now use the equation for put-call parity described above to find the solution:
P (108) - C (108) = 108 ¥ P (0,2) - 100 = 108 ¥ 0.90423 - 100 = -2.3429
Solution 40
D
Chapter 2, Option Strategies
This question is probably more appropriate for Exam FM than for Exam MFE/3F. This
material is covered on page 87 of the second edition of the Derivatives Markets textbook,
which is assigned for Exam FM but is not assigned for Exam MFE/3F.
A bull spread consists of purchasing a low-strike option and selling a high-strike option.
In the profit diagram below, the dotted line represents zero profit. The profit diagram
shows the pattern of profits at the end of 1 year:
The diagram above matches Portfolio IV, so the correct answer must be Choice (D) or
Choice (E).
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SOA Sample Exam MFE/3F
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A straddle consists of the purchase of a call and a put with the same strike price. The
profit diagram shows the pattern of the profits at the end of 1 year:
The diagram above matches Portfolio II, so the correct answer must be Choice (D).
For the sake of completeness, let’s consider the other two strategies as well.
A collar consists of the purchase of a put and the sale of a call with a higher strike price.
We would need more information to know where it would produce zero profit, so the
dotted line has been left off of the graph below. Nonetheless, the pattern indicates that
Portfolio I is the collar:
Collar
A strangle consists of the purchase of a put and the purchase of a call with a higher strike
price. It’s profit diagram matches Portfolio III.
Strangle
Solution 41
C
Chapter 8, Elasticity
The contingent claim can be replicated with a portfolio consisting of the present value of
\$42 and a short position in a European put option with a strike price of \$42.
To find the value of this contingent claim, we must find the value of the put option:
The first step is to calculate d1 and d2 :
d1 =
ln(S / K ) + (r - d + 0.5s 2 )T
s T
=
ln(45 / 42) + (0.07 - 0.03 + 0.5 ¥ 0.252 ) ¥ 1
0.25 1
= 0.56097
d2 = d1 - s T = 0.56097 - 0.25 1 = 0.31097
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SOA Sample Exam MFE/3F
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We have:
N ( d1 )  N ( 0.56097)  0.28741
N ( d2 )  N ( 0.31097)  0.37791
The value of the European put option is:
P (42)  Ke rT N ( d2 )  Se  T N ( d1 )
 42e 0.07  0.37791  45e 0.03  0.28741  2.24795
The current value of the contingent claim is the present value of \$42 minus the value of
the put:
V = 42e -0.07 - 2.24795 = 36.91259
The delta of the put option is:
D Put = - e -d T N ( -d1 ) = -e -0.03 ¥ 0.28741 = -0.27892
The delta of the contingent claim is the delta of the present value of \$42 (i.e., zero) minus
the delta of the put:
D = 0 - ( -0.27892) = 0.27892
The elasticity of the contingent claim is:
W=
S D 45 ¥ 0.27892
=
= 0.34003
V
36.91259
Solution 42
D
Chapter 10, Barrier Options
As the barrier of an up-and-out call approaches infinity, the price of the up-and-out call
approaches the price of a regular call.
Therefore, a regular call option with a strike price of \$60 can be read from the bottom line
of the table:
C (60) = 4.0861
The value of the corresponding up-and-in calls can be determined with the parity
relationship for barrier options:
Up-and-in call + Up-and-out call = Ordinary call
For the \$70 and \$80 barriers, we have:
H
Up-and-out Call
Up-and-in Call
70
0.1294
4.0861 - 0.1294 = 3.9567
80
0.7583
4.0861 - 0.7583 = 3.3278
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The special option consists of 2 of the 70-barrier up-and-in calls and a short position in 1
of the 80-barrier up-and-in calls. Therefore, the price of the special option is:
2 ¥ 3.9567 - 3.3278 = 4.5856
Solution 43
E
Chapter 14, Geometric Brownian Motion
For this question, we make use of the following relationship:
dS (t )
= (a - d )dt + s dZ (t )
S (t )
¤
2
S (t ) = S (0)e(a -d - 0.5s )t +s Z (t )
We can write x (t ) as:
2
x (t ) = x (0)e( r - r€ - 0.5s )t +s Z (t )
Therefore, y(t ) is:
y(t ) =
1
1 -( r - r€ - 0.5s 2 )t - s Z (t )
1 ( r€ - r + 0.5s 2 )t - s Z (t )
=
e
=
e
x (t ) x (0)
x (0)
This implies that:
dy(t ) È
= r€ - r + 0.5s 2 + 0.5( -s )2 ˘ dt - s dZ (t ) = (r€ - r + s 2 )dt - s dZ (t )
Î
˚
y(t )
Solution 44
D
Chapter 4, Three-Period Binomial Tree
The risk-neutral probability of an upward movement is:
p* =
(0.10 - 0.065) ¥1 210
- 300
e ( r - d )h - d e
=
= 0.6102
375 - 210
u-d
300
300
The tree of prices for the American put option is shown below:
American Put
39.7263
0.0000
14.4603
0.0000
41.0002
0.0000
116.2500
90.0000
153.0000
197.1000
Early exercise is optimal if the stock price falls to \$210 at the end of 1 year or falls to \$147
at the end of 2 years. This is indicated by the bolding of those two nodes in the tree
above.
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SOA Sample Exam MFE/3F
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The price of the option is:
e -0.10 ¥1 [0.6102 ¥ 14.4603 + (1 - 0.6102) ¥ 90] = 39.7263
Solution 45
C
Chapter 4, Greeks in the Binomial Model
We need to calculate the two possible values of delta at the end of 1 year:
Vuu  Vud
0.00  41.00
 e 0.0651
 0.1863
2
468.75  262.50
Su  Sud
V  Vdd
41.00  153.00
(Sd, h )  e  h ud
 e 0.0651
 0.9087
2
262.50  147.00
Sud  Sd
(Su, h )  e  h
We can now calculate gamma:
G(S ,0) ª G(Sh , h) =
D(Su, h ) - D(Sd, h ) -0.1863 - ( -0.9087)
=
= 0.004378
375 - 210
Su - Sd
Solution 46
E
Chapter 4, Options on Futures Contracts
We are given that the ratio of the factors applicable to the futures price is:
uF 4
=
dF 3
The formula for the risk-neutral probability of an up move can be used to find uF and
dF :
p* =
1 - dF
uF - dF
p* =
1
dF
uF
dF
d
- dF
F
d
- dF
F
1
1 dF - 1
=
4 -1
3
3
dF = 0.9
uF =
4
¥ dF = 1.2
3
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SOA Sample Exam MFE/3F
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The tree of futures prices is therefore:
Futures Prices
115.2000
96.0000
80.0000
86.4000
72.0000
64.8000
The tree of prices for the European call option is:
European Call
30.2000
10.7284
3.7838
1.4000
0.4551
0.0000
The price of the European call is:
e -0.05 ¥ 0.5 [(1 / 3)10.7284 + (2 / 3)(0.4551)] = 3.7838
The tree of prices for the American call option is:
American Call
30.2000
11.0000
3.8721
1.4000
0.4551
0.0000
The price of the American call is:
e -0.05 ¥ 0.5 [(1 / 3)11.0000 + (2 / 3)(0.4551)] = 3.8721
If the futures price moves up at the end of 6 months, then early exercise of the American
option is optimal, and this is indicated by the bolding of that node above.
The price of the American call option exceeds the price of the European call option by:
3.8721 - 3.7838 = 0.08830
Solution 47
B
Chapter 9, Market-Maker Profit
Let’s assume that “several months ago” was time 0. Further, let’s assume that the
options expire at time T and that the current time is time t. We can use put-call parity to
obtain a system of 2 equations.
8.88 + Ke -rT = 40 + 1.63
14.42 + Ke -r(T - t ) = 50 + 0.26
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SOA Sample Exam MFE/3F
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This can be solved to find ert :
Ke - rT = 32.75 ¸Ô
˝
Ke -r(T - t ) = 35.84 Ô˛
ﬁ
ert =
35.84
32.75
The market-maker sold 100 of the call options. From the market-maker’s perspective, the
value of this short position was:
-100 ¥ 8.88 = -888.00
The delta of the position, from the perspective of the market-maker, was:
Delta of a short position in 100 calls = -100 ¥ (0.794) = -79.4
To delta-hedge the position, the market-maker purchased 79.4 shares of stock. The value
of this position was:
79.4 ¥ 40 = 3,176.00
Since the market-maker received only \$888.00 from the sale of the calls, the differential
was borrowed. The value of the loan, from the perspective of the market-maker, was:
888.00  3,176.00  2,288.00
The initial position, from the perspective of the market-maker, was:
Value
–888.00
3,176.00
–2,288.00
0.00
Component
Options
Shares
Risk-Free Asset
Net
After t years elapsed, the value of the options changed by:
-100 ¥ (14.42 - 8.88) = -554.00
After t years elapsed, the value of the shares of stock changed by:
79.4 ¥ (50.00 - 40.00) = 794.00
After t years elapsed, the value of the funds that were borrowed at the risk-free rate
changed by:
(
)
Ê 35.84
ˆ
-2,288.00 ert - 1 = -2,288.00 Á
- 1˜ = -215.88
Ë 32.75
¯
The sum of these changes is the profit.
Component
Gain on Options
Gain on Stock
Interest
Overnight Profit
Change
–554.00
794.00
–215.88
24.12
The change in the value of the position is \$24.12, and this is the profit.
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Solution 48
E
Chapter 15, Sharpe Ratio
The Sharpe ratio of Asset 1 must be equal to that of Asset 2:
0.06  0.04 0.03  0.04

0.02
k
k  0.01
Over the next instant, the returns on Stock 1 and Stock 2 are:
Return on S1 = 0.06S1dt + 0.02S1dZ = 0.06(100)dt + 0.02(100)dZ = 6.0dt + 2dZ
Return on S2 = 0.03S2 dt - 0.01S2 dZ = 0.03(50)dt - 0.01(50)dZ = 1.5dt - 0.5dZ
Since the investor purchases 1 share of Stock 1, the amount of Stock 2 that must be
purchased to remove the random term (i.e., the dZ term) is:
2
=4
0.50
When the return on Stock 1 is added to the return on 4 shares of Stock 2, there is no
random term:
Return on S1  6.0dt  2dZ
Return on 4 shares of S2  4  (1.5dt  0.5dZ )
 12dt
Solution 49
B
Chapter 4, American Put Option
The values of u and d are:
u  e( r   )h   h  e(0.04  0.00)0.25  0.3 0.25  1.1735
d  e( r   )h   h  e(0.04  0.00)0.25  0.3 0.25  0.8694
The risk-free probability of an upward movement is:
p* =
e( r - d )h - d e(0.04 - 0.00)0.25 - 0.8694
=
= 0.4626
u-d
1.1736 - 0.8694
The stock price tree is:
117.35
100
86.94
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The strike price K, for which an investor will exercise the put option at the beginning of
the period must be at least \$100, since otherwise the payoff to immediate exercise would
be zero. Since we are seeking the lowest strike price that results in immediate exercise,
let’s begin by determining whether there is a strike price that is greater than \$100 but
less than \$117.35 that results in immediate exercise. If there is such a strike price, the
value of exercising now must exceed the value of holding the option:
K  100  ( K  86.94)(1  p*)e 0.04(0.25)
K  100  ( K  86.94)(1  0.4626)e 0.04(0.25)
K  100  0.5321K  46.26
0.4679 K  53.74
K  114.85
The smallest integer that satisfies this inequality is \$115.
Solution 50
A
Chapter 5, Lognormal Confidence Intervals
We are given:
S0 = 0.25
s = 0.35
a - d = 0.15
The z-value associated with the upper limit of the 90% confidence interval is found below:
(
P (z > z
P (z > z
P (z < z
) 2p
) = 0.10
2
) = 0.05
) = 0.95
P z > zU =
U
U
U
zU = 1.64485
The upper limit of the confidence interval is calculated using the associated z-value:
2
U
U
ST
= St e(a -d - 0.5s )(T -t ) +s z T -t
2
= 0.25e(0.15 - 0.5 ¥ 0.35 )(0.5) + 0.35 ¥1.64485 0.5 = 0.39265
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Solution 51
E
Chapter 13, Estimating Parameters of a Lognormal Distribution
First, we determine the monthly mean and the monthly standard deviation using a
calculator, and then we determine the annualized standard deviation and the annualized
expected return.
Using the TI-30X IIS calculator, press [2nd] [STAT] and choose 1-VAR and press
[ENTER] [DATA]. Perform the following sequence:
X1 = LN(56/54) [ENTER] ↓↓ (Hit the down arrow twice)
X2 = LN(48/56) [ENTER] ↓↓
X3 = LN(55/48) [ENTER] ↓↓
X4 = LN(60/55) [ENTER] ↓↓
X5 = LN(58/60) [ENTER] ↓↓
X6 = LN(62/58) [ENTER]
Press [STATVAR] and → and the monthly mean is shown as 0.02302506. Press → again
and the monthly standard deviation is shown as 0.103541414. Multiply this number by
the square root of 12 to get the annualized standard deviation of 0.35867798. To exit the
statistics mode, press [2nd] [EXITSTAT] [ENTER].
Since the dividend rate is zero, we determine the annualized expected return by
multiplying the monthly mean return by 12 and adding to that amount one half of the
annualized variance:
ˆ 
r
   0.5ˆ 2  0.02302506  12  0  0.5  0.358677982  0.340625623
h
Solution 52
C
Chapter 12, Normal Random Variables as Quantiles
We convert the draws from the uniform distribution into draws from the standard normal
distribution:
F (0.98300)  N ( zˆ1 ) 
F (0.03836)  N ( zˆ2 ) 
F (0.77935)  N ( zˆ3 ) 
zˆ1  2.12007
zˆ2  1.77004
zˆ3  0.77000
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We use these draws to find the new stock prices:
2
ST  St e(    0.5 )(T  t )   z T  t
2
1. S2  50e(0.15  0.00  0.5 0.3 )2  0.30 (2.12007) 2  151.63746
2
2. S2  50e(0.15  0.00  0.5 0.3 )2  0.30 ( 1.77004) 2  29.10933
2
3. S2  50e(0.15  0.00  0.5 0.3 )2  0.30 (0.77000) 2  85.51624
The average of the stock prices is:
151.63746 + 29.10933 + 85.51624
= 88.75434
3
Solution 53
B
Chapter 11, Cash-Or-Nothing Call Option
A European call option has the same gamma as an otherwise equivalent European put
option. Therefore, the gamma of a call option with the payoff described below is 0.07:
ST - 40
if ST > 40
The gap call option described in the question has a gamma of 0.08, and its payoff is:
ST - 45
if ST > 40
The cash-or-nothing call option has the following payoff:
1,000
if ST  40
The cash-or-nothing call option can be replicated by purchasing 200 of the call options and
selling 200 of the gap call options:
200 [Call - GapCall ] = 200 [(ST - 40) - (ST - 45)] = 200 ¥ 5 = 1,000
if ST > 40
The gamma of the position is:
200 [0.07 - 0.08] = 200 ¥ ( -0.01) = -2.00
Solution 54
A
Chapter 1, Exchange Options
Consider an asset, which we will call Asset X, which has a payoff of:
X1  Min[2S1 (1), S2 (1)]
From statement (vi), we know that the price of a European call option on Asset X, with a
strike price of 17, is 1.632. The payoff of this call option is:
Max Min 2S1 (1), S2 (1)  17,0  Max X1  17,0
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We are asked to find the price of an option that has the payoff described below:
Max 17  Min 2S1 (1), S2 (1) ,0  Max 17  X1 ,0
We recognize this as a European put option on Asset X. From put-call parity, we have:
CEur + Ke -r(T - t ) = St + PEur
1.632 + 17e -0.05 ¥1 = X 0 + PEur
If we can find the current value of Asset X, then we can solve for the value of the put
option. Asset X can be replicated by purchasing 2 shares of Stock 1, and selling an
exchange call option that allows its owner to exchange 1 share of Stock 2 for 2 shares of
Stock 1:
X1  Min[2S1 (1), S2 (1)]  2S1 (1)  Max[2S1 (1)  S2 (1),0]
The current value of Asset X is:
X 0  2S1 (0)  ExchangeCallPrice
The exchange call option is an exchange call option with 2 shares of Stock 1 as the
underlying asset and 1 share of Stock 2 as the strike asset. To find the value of this
exchange call option, we find the appropriate volatility parameter:
2
s = s S2 + s K
- 2 rs S s K = 0.182 + 0.252 - 2( -0.40)(0.18)(0.25) = 0.36180
The values of d1 and d2 are:
d1 =
Ê Se -d ST ˆ s 2T
ln Á
˜+
2
Ë Ke -d K T ¯
s T
=
Ê (2 ¥ 10)e -0 ¥1 ˆ 0.361802 ¥ 1
ln Á
˜+
2
Ë 20e -0 ¥1 ¯
0.36180 1
= 0.18090
d2 = d1 - s T = 0.18090 - 0.36180 1 = -0.18090
From the normal table, we have:
N ( d1 ) = N (0.18090) = 0.57178
N ( d2 ) = N ( -0.18090) = 0.42822
The value of the exchange call option is:
ExchangeCallPrice  Se  ST N (d1 )  Ke  K T N (d2 )
 2  10e0 1 (0.57178)  20e0 1 (0.42822)
 2.8712
The current value of Asset X is:
X 0  2S1 (0)  ExchangeCallPrice  2  10  2.8717  17.1288
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We can now use put-call parity to find the value of the put option:
1.632 + 17e -0.05 ¥1 = X 0 + PEur
1.632 + 17e -0.05 = 17.1288 + PEur
PEur = 0.6741
Solution 55
D
Chapter 7, Options on Futures Contracts
At time 0, the values of d1 and d2 are:
d1 


ln F0,TF / K  0.5 2T
 T

ln  20 / 20   0.5 2T
 T
d2  d1   T  0.5 T   T  0.5 T
 0.5 T
We can find N ( -d1 ) in terms of N ( -d2 ) :
(
)
N ( -d2 ) = N (0.5s T )
N ( -d1 ) = N -0.5s T
N ( -d1 ) = 1 - N ( -d2 )
We can substitute this value of N ( -d1 ) into the time 0 formula for the price in order to
find d2 :
PEur ( F0,TF , K , , r ,T , r )  Ke rT N ( d2 )  F0,TF e rT N ( d1 )
1.625  20e 0.10  0.75 N ( d2 )  20e 0.10  0.75 N ( d1 )
1.625  20e 0.075  N ( d2 )  1  N ( d2 )  
1.625 0.075
 2N ( d2 )  1
e
20
N ( d2 )  0.54379
d2  0.10999
d2  0.10999
We can now find s :
d2  0.5 T
0.10999  0.5 0.75
  0.25401
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After 3 months, the new values of d1 and d2 are:
d1 


ln F0,TF / K  0.5 2T
 T

ln 17.7 / 20   0.5  0.254012  0.5
0.25401 0.5
 0.59037
d2  d1   T  0.59037  0.25401 0.5  0.76998
We can look up the values of N ( -d1 ) and N ( -d2 ) from the normal distribution table:
N ( -d1 ) = N (0.59037) = 0.72253
N ( -d2 ) = N (0.76998) = 0.77934
After 3 months, the value of the put option is:
PEur ( F0,TF , K ,s , r ,T , r ) = Ke - rT N ( -d2 ) - F0,TF e -rT N ( -d1 )
= 20e -0.10 ¥ 0.5 ¥ 0.77934 - 17.70e -0.10 ¥ 0.5 ¥ 0.72253
= 2.66156
Solution 56
A
Chapter 5, Covariance of St and ST
To answer this question, we use the following formulas:
ÈS ˘
E Í T ˙ = e(a - d )(T - t )
Î St ˚
2
Var ÈÎST St ˘˚ = St2 e2(a - d )(T - t ) Ê es (T - t ) - 1ˆ
Ë
¯
ÈS
Cov[St , ST ] = E Í T
Î St
˘
˙ Var ÈÎSt S0 ˘˚
˚
The variance of A(2) is:
Var [ A(2)] =
1
{Var[S(1) + Var[S(2)] + 2Cov[S(1), S(2)]}
4
The variances of S (1) and S (2) are:
2
Var ÈÎS1 S0 ˘˚ = 52 e2(0.05)(1 - 0) Ê e0.2 (1 - 0) - 1ˆ = 1.12757
Ë
¯
2
Var ÈÎS2 S0 ˘˚ = 52 e2(0.05)(2 - 0) Ê e0.2 (2 - 0) - 1ˆ = 2.54318
Ë
¯
The covariance is:
ÈS ˘
Cov[S1 , S2 ] = E Í 2 ˙ Var ÈÎS1 S0 ˘˚ = e0.05(2 -1) ¥ 1.12757 = 1.18538
Î S1 ˚
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We can now find the variance of A(2) :
1
{Var[S(1) + Var[S(2)] + 2Cov[S(1), S(2)]}
4
1
= {1.12757 + 2.54318 + 2 ¥ 1.18538} = 1.51038
4
Var [ A(2)] =
Solution 57
E
Chapter 12, Stratified Sampling Method
The stratified sampling method assigns the first and fifth uniform (0, 1) random numbers
to the segment (0.00, 0.25), the second and sixth uniform (0, 1) random variables to the
segment (0.25, 0.50), the third and seventh uniform (0, 1) random variables to the
segment (0.50, 0.75), and the fourth and the eighth uniform (0, 1) random variables to the
segment (0.75, 1.00).
The lowest simulated normal random variables will come from the segment (0.00, 0.25).
The lower of the two values in this segment is the fifth one: 0.3172, so we use it to find the
corresponding standard normal random variable:
u5 = 0.3172
uˆ5 =
0.3172 + (1 - 1)
= 0.0793
4
Z5 = N -1 (0.0793) = -1.40980
The highest simulated normal random variable will come from the segment (0.75, 1.00).
The higher of the two values in this segment is the fourth one: 0.4482, so we use it to find
the corresponding standard normal random variable:
u4 = 0.4482
uˆ 4 =
0.4482 + (4 - 1)
= 0.86205
4
Z4 = N -1 (0.86205) = 1.08958
The difference between the largest and smallest simulated normal random variates is:
1.08958 - ( -1.40980) = 2.49938
Solution 58
B
Chapter 12, Control Variable Method with Beta
Let the variable Y be associated with the \$42-strike option and the variable X be
associated with the \$40-strike option. In the regression analysis, Y will be the dependent
variable and X will be the independent variable.
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55
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Solutions
The payoffs depend on the simulated stock prices:
Payoff
With
Strike = 40
Stock
Price
(X e )
33.29
37.30
40.35
43.65
48.90
0.00
0.00
0.35
3.65
8.90
i
Payoff
With
Strike = 42
(Y e )
0.02
i
0.02
0.00
0.00
0.00
1.65
6.90
The values in the two rightmost columns are the payoffs, which means that they are the
discounted Monte Carlo prices, X i and Yi , times erT .
The estimate for b is:
n
b =
Â (Yi - Y )( Xi - X )
i =1
n
Â
=
2 n
(e )
rT
Â (Yi - Y )( Xi - X )
i =1
i =1
2 n
( )
erT
( X i - X )2
Â ( Xi - X )2
i =1
We get the same estimate for b regardless of whether we use the time 0 prices or the
time 0.25 payoffs (as shown in the rightmost expression above). To save time, we use the
time 0.25 payoffs from the table above.
We perform a regression using the 2nd column in the table above as the x-values and the
third column as the y-values. The resulting slope coefficient is:
b = 0.764211
During the exam, it is more efficient to let the calculator perform the regression and
determine the slope coefficient.
Using the TI-30XS Multiview calculator, we first clear the data by pressing [data] [data]
↓↓↓ until Clear ALL is shown, and then press [enter].
Fill out the table as shown below:
L1
0.00
0.00
0.35
3.65
8.90
L2
0.00
0.00
0.00
1.65
6.90
L3
-----------
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56
SOA Sample Exam MFE/3F
Solutions
Next, press:
[2nd] [stat] 2 (i.e., select 2-Var Stats)
Select L1 for x-data and press [enter].
Select L2 for y-data and press [enter].
Select CALC and then [enter]
Scroll down to: a = 0.764211005.
Exit the statistics mode by pressing [2nd] [quit].
Solution 59
B
Chapter 12, Control Variable Method with Beta
Since the 40-strike call is our control, we have:
0 + 0 + 0.35 + 3.65 + 8.90
¥ e -0.02 = 2.5289
5
0 + 0 + 0 + 1.65 + 6.90
¥ e -0.02 = 1.6761
Y =
5
X =
The estimate for the price of the \$42-strike option is:
C * (42)  Y *  Y   ( X  X )  1.6761  0.764211(2.7847  2.5289)  1.8716
Solution 60
E
Chapter 19, Cox-Ingersoll-Ross Model
The CIR model is:
dr = a(b - r )dt + s rdZ
From the partial differential equation provided in the question, we can obtain the
following parameters for the CIR model:
a = 0.1
b = 0.11
s = 0.08
We can define c in terms of f :
f (r, t ) = f
r
s
&
f (r, t ) = c r
ﬁ
c=
f
s
In the CIR model, as the maturity of a zero-coupon bond approaches infinity, its yield
approaches:
r 
2ab
  ln[ P (r, t,T )] 
 Lim 

T t
(a     ) T  

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From the information provided in part (ii) of the question, we know that the rightmost
portion is equal to 0.10. Therefore:
2ab
 0.10
(a     )
2  0.1  0.11
 0.10
(0.1     )
0.22  0.1    
    0.12
We can substitute this value into the formula for g :
g = (a - f )2 + 2s 2
f + 0.12 = (0.1 - f )2 + 2(0.08)2
f 2 + 0.24f + 0.0144 = (0.1 - f )2 + 2(0.08)2
f 2 + 0.24f + 0.0144 = (0.01 - 0.2f + f 2 ) + 0.0128
0.44f = 0.0084
f = 0.01909
The value of c is:
c
 0.01909

 0.2386

0.08
Solution 61
D
Chapter 15, Risk-Neutral Process
In the Black-Scholes framework:
dS (t )
= (a - d )dt + s dZ (t )
S (t )
We can use statements (ii) and (iii) in the question to find a and s :
a - d = 0.05
s = 0.25
ﬁ
a - 0.01 = 0.05
ﬁ
a = 0.06
When the stock price process is a geometric Brownian motion, the relationship between
Z (T ) and Z (T ) is described by:
a -r
Z (T ) = Z (T ) +
T
s
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Solutions
Under the risk-neutral probability measure Z (T ) is a pure Brownian motion, so:
a -r ˘
È
E * ÈÎ Z (T ) ˘˚ = E * Í Z (T ) +
T˙
s
Î
˚
a -r
T
0 = E * [Z (T )] +
E * [Z (T )] =
r -a
s
s
T
Since the expectation of Z (0.5) is -0.03 , we have:
E * [Z (0.5)] =
r -a
s
¥ 0.5
r - 0.06
¥ 0.5
0.25
r = 0.045
-0.03 =
Solution 62
A
Chapter 15, Risk-Neutral Process
The risk-neutral price process is:
dS (t )
= (r - d )dt + s dZ (t )
S (t )
We observe that:
m = r -d
and s = 0.4
The formula for the forward price of S 2 is:
a( r  )  0.5a( a 1) 2  (T t )

a
Ft ,T S (T )a   S (t ) e 


2
2

2 2   0.52(2 1)0.40  (T t )
e
 S (t )
2  0.16(T t )
 S (t ) e
Setting this expression equal to the forward price provided in statement (ii), we have:
S(t )2 e2  0.16(T t )  S(t )2 e0.18(T t )
2  0.16  0.18
  0.01
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Solution 63
A
The quadratic variation is the sum of the squared increments of the process.
(i)
W is not stochastic, and its differential is found below:
W (t ) = t 2
dW = 2tdt
Making use of the multiplication rule, dt ¥ dt = 0 , we obtain the quadratic
variation:
2
V2.4
(W )
2.4
=
2.4
2
Ú [dW ]
=
0
(ii)
2
Ú [2tdt ]
0
2.4
=
Ú 4t
2
(dt ¥ dt )] = 0
0
The increments to X are 0 across the interval [0,1). But when t = 1 , X jumps from
0 to 1. The subsequent increments are again 0 until t = 2 , at which point X jumps
from 1 to 2. The subsequent increments, up to t = 2.4 , are 0. An infinite number
of zeros is summed, and their sum is zero. The two non-zero increments (of 1 each)
sum to 2:
2
V2.4
(X )
2.4
=
2
Ú [dX ]
= 02 +  + 02 + 12 + 02 +  + 02 + 12 + 02 +  + 02 = 2
0
(iii)
We can determine dY by finding the partial differential of each of the terms of Y:
Y = 2t + 0.9Z
dY = 2dt + 0.9dZ
Making use of the multiplication rules, we have:
(dY )2 = (2dt + 0.9dZ )2 = 0.81(dZ )2 = 0.81dt
2
V2.4
(Y )
2.4
=
2
Ú [dY ]
2.4
=
0
Ú 0.81dt = 0.81 ¥ (2.4 - 0) = 1.944
0
Since 0 < 1.944 < 2 , we have:
2
2
2
V2.4
(W ) < V2.4
(Y ) < V2.4
(X )
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60
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Solution 64
C
Chapters 15, Prediction Intervals
Y (t ) follows geometric Brownian motion with the following parameters:
aY - dY = 1.2
s Y = -0.50
s Y = s Y = -0.50 = 0.50
The upper z-value for the 90% prediction interval is found below:
(
)
P z > zU =
p
2
(
)
ﬁ P z > zU =
1 - 0.90
2
(
)
ﬁ P z < zU = 0.95 ﬁ zU = 1.64485
The upper limit of the prediction interval for Y (2) is:
2
U
Y (0)e(aY -dY - 0.5s Y )T +s Y z
T
2
= 64e(1.2 - 0.5 ¥ 0.50 )(2) + 0.50 ¥1.64485 ¥ 2 = 1,758.0626
The corresponding upper limit of the prediction interval for S (2) is the square root of the
upper limit for Y (2) :
U = 1,758.0626 = 41.9293
Solution 65
C
Chapter 8, Elasticity and Risk Premium
Since the Black-Scholes framework applies, we have:
ÊS ˆ
ln Á T ˜  N È(a - d - 0.5s 2 )(T - t ), s 2 (T - t ) ˘
Î
˚
Ë St ¯
Therefore, from (iii), we have:
a - d - 0.5s 2 = 0.10
Under the risk-neutral probability measure, the expected return is r, and so from (iv), we
have:
(r - d - 0.5s 2 )(5 - 3) = 0.06
r - d - 0.5s 2 = 0.03
Subtracting the expression above from the one obtained from (iii), we have the risk
a - r = 0.10 - 0.03 = 0.07
In (vii), we are not told if the put option being hedged has been purchased or sold.
Therefore, we know that one of the following is true:
S D = 20
or
- S D = 20
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Since the D of a put is negative, the second expression above must be true. Therefore, we
have:
S D = -20
We can now find the elasticity of the put option:
W=
S D -20
=
= -2
V
10
The formula for the risk premium of the option allows us to solve for the absolute value of
its expected return:
g - r = W ¥ (a - r )
g - 0.04 = -2 ¥ (0.07)
g = -0.10
g = 0.10
Solution 66
A
Chapter 15, Sharpe Ratio
From (i), we have:
d X = 0.02
dY = 0.01
From (ii), we have:
a X - d X = 0.06
s X = 0.20
ﬁ
a X = 0.06 + d X = 0.06 + 0.02 = 0.08
From (iii), we have:
m = aY - dY - 0.5s Y2
s Y = -0.10
Stocks X and Y must have the same Sharpe ratio:
a X - r aY - r
=
sX
sY
0.08 - 0.04 aY - 0.04
0.20
aY = 0.02
=
-0.10
We can now find m :
m = aY - dY - 0.5s Y2 = 0.02 - 0.01 - 0.5 ¥ ( -0.10)2 = 0.005
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Solution 67
A
Chapter 15, Sharpe Ratio
Stock 1 does not pay dividends, so its parameters are:
a1 = m
s1 = 20 m
and
Stock 2 has a dividend yield of 0.01, so its parameters are:
a 2 - d 2 - 0.5s 22 = 0.03
s 2 = 0.20
a 2 - 0.01 - 0.5 ¥ (0.20)2 = 0.03
a 2 = 0.06
Since the two stocks have the same source of uncertainty, they must have the same
Sharpe ratio:
a1 - r a 2 - r
=
s1
s2
m - 0.04 0.06 - 0.04
=
20 m
0.20
m - 0.04 = 2 m
m = -0.04
Solution 68
D
Chapter 14, Ornstein-Uhlenbeck Process
The key to this question is recognizing that the process is an Ornstein-Uhlenbeck process:
dX (t ) = l ¥ [a - X (t )]dt + s dZ (t )
¤
(
)
t
X (t ) = X (0)e - l t + a 1 - e - l t + s e - l (t - s ) dZ ( s)
Ú
0
The parameters of the process are:
l =3
a =0
s =2
Therefore the solution is:
t
t
È
˘
X (t ) = X (0)e -3t + 0 ¥ 1 - e -3t + 2 e -3(t - s ) dZ ( s) = e -3t Í X (0) + 2 e3s dZ ( s) ˙
Í
˙
0
0
Î
˚
(
)
Ú
Ú
In this form, we can observe that:
A =3
B = X (0)
C=2
D=3
The sum of A, C, and D is:
A +C + D = 3+2+3 = 8
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Solution 69
C
Chapter 4, Theta in the Binomial Model
The formula for theta is:
q (S ,0) =
Vud - V - (Sud - S ) D(S ,0) -
(Sud - S )2
G(S ,0)
2
2h
Since we have S = Sud = 120 , this simplifies to:
q (S ,0) =
Vud - V
2h
The risk-neutral probability of an upward move is:
p* =
120e0.10 - 96
= 0.67816
150 - 96
Since the put option is an American option, we solve for its value from right to left. The
completed tree is shown below.
0.0000
8.2381
2.0353
6.9892
24.0000
0.0000
0.0000
24.0000
43.2000
58.5600
The bolded entries in the table above indicate where early exercise is optimal.
The value of theta can now be found:
q (S ,0) =
Vud - V 6.9892 - 8.2381
=
= -0.6245
2h
2 ¥1
Solution 70
C
Chapter 14, Portfolio Returns
The instantaneous percentage increase of the fund is the weighted average of the return
on the stock (including its dividend yield) and the return on the risk-free asset:
dW (t )
È dS (t )
˘
= 0.80 Í
+ d dt ˙ + 0.20rdt
W (t )
Î S (t )
˚
= 0.80 [0.10dt + 0.20dZ (t ) + 0.02dt ] + 0.20(0.05)dt
= 0.106dt + 0.16dZ (t )
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Since the fund’s value follows geometric Brownian motion, we can convert the partial
differential equation into an expression for the value of the fund:
dW (t )
= (aW - dW )dt + s W dZ (t )
W (t )
dW (t )
= 0.106dt + 0.16dZ (t )
W (t )
2
W(t ) = W (0)e(aW -dW - 0.5s W )t +s W Z (t )
¤
2
W(t ) = W (0)e(0.106 - 0.5 ¥ 0.16 )t + 0.16Z (t )
¤
Simplifying the lower right-hand expression above, we have:
W(t ) = W (0)e0.0932t + 0.16Z (t )
Solution 71
C
Chapter 15, Risk-Neutral Pricing
The risk-neutral price process is:
dS (t )
= (r - d )dt + s dZ (t )
S (t )
From statement (iii), we can obtain the risk-free rate and the volatility parameter:
r - d = 0.08
ﬁ
r = 0.08 + 0.04 = 0.12
s = 0.40
We can use the following equivalency:
dS (t )
= (r - d )dt + s dZ (t )
S (t )
dS (t )
= 0.08dt + 0.40dZ (t )
S (t )
¤
2

S (t ) = S (0)e( r -d - 0.5s )t +s Z (t )
¤
2


S (t ) = S (0)e(0.08 - 0.5 ¥ 0.4 )t + 0.4 Z (t ) = e0.4 Z (t )
Let’s find the expected value of the derivative security under the risk-neutral probability
measure:
{
}
2˘
È


2
E * ÍÈ1 + S (1) ¥ ln [S (1)] ˙˘ = E * Í1 + e0.4Z (1) ¥ ln È e0.4 Z (1) ˘ ˙
ÍÎ
˙˚ ˙
Î
˚
ÎÍ
˚
{
}
{
}
{ }
2˘
2˘


È
È
= E * Í1 + e0.4 Z (1) ¥ 0.4Z (1) ˙ = 1 + E * Í 0.16e0.4 Z (1) ¥ Z (1) ˙
Î
˚
Î
˚
2
 ˘
È
= 1 + 0.16E * Í Z (1) ¥ e0.4 Z (1) ˙
Î
˚
{ }
Since Z (1) is a standard normal random variable, we can use statement (v) to obtain:
{ }
(
)
2
2
 ˘
È
1 + 0.16 E * Í Z (1) ¥ e0.4 Z (t ) ˙ = 1 + 0.16 ÈÍ 1 + 0.42 e0.4 / 2 ˘˙ = 1.20106
Î
˚
Î
˚
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65
SOA Sample Exam MFE/3F
Solutions
The time-0 price is obtained by discounting the expected value at the risk-free rate:
1.20106 ¥ e - r = 1.20106 ¥ e -0.12 = 1.06524
Solution 72
A
Chapter 15, Gap Put-Call Parity and S a
The usual put-call parity expression for gap calls and gap puts is:
GapCall + K1 e - rT = Se -d T + GapPut
The first term on the right side of the equation is the prepaid forward price of the
underlying asset, so we can also write the put-call parity expression as:
GapCall + K1e -rT = F0,PT (S ) + GapPut
In this case, the underlying asset is S 2 , so we replace S with S 2 :
GapCall + K1e - rT = F0,PT (S (T )2 ) + GapPut
2
a Èa( r -d ) + 0.5a( a -1)s ˚˘T
GapCall + K1e - rT = e -rT [S (0)] e Î
+ GapPut
2
5.542 + 95e -0.07(0.5) = e -0.07(0.5) ¥ 102 e[2(0.07 -d ) + 0.5 ¥ 2(2 -1)0.10 ]0.5 - 4.745
97.2745 = e -0.035 ¥ 100e[0.15 - 2d ]0.5 - 4.745
105.6534 = 100e(0.075 -d )
d = 0.02
Solution 73
B
Chapter 15, Claim on S a
The usual relationship is:
dS (t )
= (a - d )dt + s dZ (t )
S (t )
ﬁ
d(S a )
S
a
= È a(a - d ) + 0.5a(a - 1)s 2 ˘ dt + as dZ (t )
Î
˚
In this case, a - d = 0.30 and s = -s , so we have:
dS (t )
= 0.30dt - s dZ (t )
S (t )
ﬁ
d(S a )
S
a
= È0.30a + 0.5a(a - 1)s 2 ˘ dt - as dZ (t )
Î
˚
From statement (ii), we have the following 2 equations and 2 unknowns:
ÏÔ0.30a + 0.5a(a - 1)s 2 = -0.66
Ì
ÔÓ -as = 0.6
Page
66
SOA Sample Exam MFE/3F
The second equation can be written as a =
Solutions
-0.6
s
, and we can substitute this value of a
into the first equation:
0.30a + 0.5a(a - 1)s 2 = -0.66
Ê -0.6 ˆ
Ê -0.6 ˆ È -0.6
˘
0.30 Á
+ 0.5 Á
- 1˙ s 2 = -0.66
˜
˜
Í
Ë s ¯
Ë s ¯Î s
˚
-0.18
+ 0.18 + 0.3s = -0.66
s
-0.18
s
+ 0.84 + 0.3s = 0
0.3s 2 + 0.84s - 0.18 = 0
s 2 + 2.8s - 0.6 = 0
10s 2 + 28s
(10s - 2)(s
10s - 2 = 0
s = 0.2 or
-6=0
+ 3) = 0
or s + 3 = 0
s = -3
Since we are told that s is a positive constant, we conclude that s = 0.2 .
Solution 74
E
Chapter 5, Probability that Stock Price is Greater Than K
Let F be the fund amount. The time-4 cash flow to the company that sold the put option
is:
F (1.02)4 - Max [0,40 - S4 ]
For this cash flow to be negative, the put option must be in-the-money, meaning that S4
must be less than 40. Furthermore, the put option’s nonzero payoff must be greater than
F (1.02)4 . That is, the following must be negative:
F (1.02)4 - (40 - S4 )
The company wants the probability of a nonnegative cash flow to be 99%, so:
P È F (1.02)4 - (40 - S4 ) > 0 ˘ = 0.99
Î
˚
P È F (1.02)4 - 40 + S4 > 0 ˘ = 0.99
Î
˚
P È S4 > 40 - F (1.02)4 ˘ = 0.99
Î
˚
Page
67
SOA Sample Exam MFE/3F
Solutions
We make use of the following formula for the probability of the price of a stock exceeding
a threshold:
Prob (ST > K ) = N (dˆ2 )
with
K = 40 - F (1.02)4
We can use the normal distribution table to find dˆ2 :
N ( dˆ2 ) = 0.99
dˆ2 = 2.32635
ﬁ
We have:
dˆ2 =
ÊS ˆ
ln Á t ˜ + (a - d - 0.5s 2 )(T - t )
ËK¯
2.32635 =
s T -t
Ê
ˆ
40
ln Á
+ (0.10 - 0 - 0.5 ¥ 0.302 )(4 - 0)
4˜
Ë 40 - F (1.02) ¯
0.30 4 - 0
Ê
ˆ
40
1.17581 = ln Á
˜
Ë 40 - F (1.02)4 ¯
F = 25.55102
Solution 75
D
Chapter 12, Variance of Control Variate Estimate
The formula from the ActuarialBrew.com Study Manual that has X as the control variate
and Y * as the control variate estimate is:
(
Var [Y * ] = Var ÈÎY ˘˚ 1 - r 2
X ,Y
)
In this question, however, the control variate is denoted by Y and the control variate
estimate is denoted by X * , so we switch the roles of X and Y:
(
Var [ X * ] = Var ÈÎ X ˘˚ 1 - r 2
X ,Y
) = 5 (1 - 0.8 ) = 9
2
2
Solution 76
D
Chapter 18, BDT Interest Rate Cap
The ratio of each interest rate to the rate above it in the tree is constant for any column
within the tree, so:
rud ruu
=
rdd rud
ﬁ
rud = ruu ¥ rdd = 0.172 ¥ 0.106 = 0.1350
Page
68
SOA Sample Exam MFE/3F
Solutions
The completed tree of interest rates is:
0.1720
0.1260
0.0900
0.1350
0.0930
0.1060
Although the cap payments are made at the end of each year, we will value them at the
beginning of each year using the following formula:
T -year caplet payoff at time (T - 1) =
Max ÈÎ 0, RT -1 - K R ˘˚
1 + RT -1
¥ Notional
A 3-year cap consists of a 1-year caplet, a 2-year caplet, and a 3-year caplet.
The payoff for the 1-year caplet is zero since 9% is less than 11.5%. Therefore, the value
of the 1-year caplet is zero.
The payoff for the 2-year caplet is positive only when the short-term interest rate
increases to 12.6%:
R1 = 0.126
ﬁ
0.126 - 0.115
¥ 10,000 = 97.6909
1.126
The payoff for the 3-year caplet is positive when the short-term rate is 17.2% or 13.5%:
R2  0.172

R2  0.135

0.172  0.115
 10,000  486.3481
1.172
0.135  0.115
 10,000  176.4358
1.135
The possible payoffs from the cap are illustrated in the tree below:
486.3481
97.6909
176.4358
0.0000
0.0000
0.0000
The value of the 2-year caplet is:
Value of 2-year caplet  0.5 
97.6909
 44.8124
1.09
The value of the 3-year caplet is:
Value of the 3-year caplet
486.3481
176.4358
176.4358
 (0.5)2 
 (0.5)2 
 (0.5)2 
 172.0279
1.09  1.126
1.09  1.126
1.09  1.093
Page
69
SOA Sample Exam MFE/3F
Solutions
The value of the 3-year cap is:
(1-year caplet)  (2-year caplet)  (3-year caplet)  0.000  44.8214  172.0279
 216.8402
Page
70
``` # HOW TO BUILD THE SUSTAINABLE SOA SOA Enablement Solution • # HOW TO GET REAL BUSINESS VALUE FROM SOA SOA Solution Brief • # SOA Patterns: New Insights or Recycled Knowledge? Gregor Hohpe # Why Is EA So Important? Jeff Tash, ITscout Flashmap Systems, Inc. # BIJIT - BVICAM’s International Journal of Information Technology 